The document provides an overview of using integrals to calculate the areas between curves, volumes of solids, lengths of curves, and other quantities. It discusses:
- Approximating the area between curves using Riemann sums and defining the integral as the limit of these sums.
- Calculating areas between curves by integrating f(x) - g(x) from the limits of integration, which may be where the curves intersect.
- Partitioning a region into subregions if the bounding curve formulas change, and integrating over each subregion.
- Calculating areas by integrating with respect to y instead of x if the region is described in terms of y.
- Combining calculus
THE QUALITATIVE ANALYZE OF THE CARTOGRAPHICAL PROJECTIONS USED IN ROMANIAmihai_herbei
In Romania there were used along the time the following projection systems: Conic projection Bonne, on the Clarke ellipsoid, between1916-1930, Stereographic projection on the Hayford ellipsoid, between 930-1951, Gauss-Kruger projection, on the Krasowsky ellipsoid, between1951-1970, Stereographic projection 1970, on the Krasowsky ellipsoid, starting with 1971. For the military applications in Romania it is used the UTM projection (Universal Transverse Mercator).
This presentation is a part of Computer Oriented Numerical Method . Newton-Cotes formulas are an extremely useful and straightforward family of numerical integration techniques.
THE QUALITATIVE ANALYZE OF THE CARTOGRAPHICAL PROJECTIONS USED IN ROMANIAmihai_herbei
In Romania there were used along the time the following projection systems: Conic projection Bonne, on the Clarke ellipsoid, between1916-1930, Stereographic projection on the Hayford ellipsoid, between 930-1951, Gauss-Kruger projection, on the Krasowsky ellipsoid, between1951-1970, Stereographic projection 1970, on the Krasowsky ellipsoid, starting with 1971. For the military applications in Romania it is used the UTM projection (Universal Transverse Mercator).
This presentation is a part of Computer Oriented Numerical Method . Newton-Cotes formulas are an extremely useful and straightforward family of numerical integration techniques.
Master Thesis on the Mathematial Analysis of Neural NetworksAlina Leidinger
Master Thesis submitted on June 15, 2019 at TUM's chair of Applied Numerical Analysis (M15) at the Mathematics Department.The project was supervised by Prof. Dr. Massimo Fornasier. The thesis took a detailed look at the existing mathematical analysis of neural networks focusing on 3 key aspects: Modern and classical results in approximation theory, robustness and Scattering Networks introduced by Mallat, as well as unique identification of neural network weights. See also the one page summary available on Slideshare.
Dumitru Vulcanov - Numerical simulations with Ricci flow, an overview and cos...SEENET-MTP
Lecture by prof. dr Dumitru Vulcanov (dean of the Faculty of Physics, West University of Timisoara, Romania) on October 21, 2010 at the Faculty of Science and Mathematics, Nis, Serbia.
Formulas for Surface Weighted Numbers on Graphijtsrd
The boundary value problem differential operator on the graph of a specific structure is discussed in this article. The graph has degree 1 vertices and edges that are linked at one common vertex. The differential operator expression with real valued potentials, the Dirichlet boundary conditions, and the conventional matching requirements define the boundary value issue. There are a finite number of eig nv lu s in this problem.The residues of the diagonal elements of the Weyl matrix in the eigenvalues are referred to as weight numbers. The ig nv lu s are monomorphic functions with simple poles.The weight numbers under consideration generalize the weight numbers of differential operators on a finite interval, which are equal to the reciprocals of the squared norms of eigenfunctions. These numbers, along with the eig nv lu s, serve as spectral data for unique operator reconstruction. The contour integration is used to obtain formulas for surfacethe weight numbers, as well as formulas for the sums in the case of superficial near ig nv lu s. On the graphs, the formulas can be utilized to analyze inverse spectral problems. Ghulam Hazrat Aimal Rasa "Formulas for Surface Weighted Numbers on Graph" Published in International Journal of Trend in Scientific Research and Development (ijtsrd), ISSN: 2456-6470, Volume-6 | Issue-3 , April 2022, URL: https://www.ijtsrd.com/papers/ijtsrd49573.pdf Paper URL: https://www.ijtsrd.com/mathemetics/calculus/49573/formulas-for-surface-weighted-numbers-on-graph/ghulam-hazrat-aimal-rasa
Parallel Evaluation of Multi-Semi-JoinsJonny Daenen
Presentation given on VLDB 2016: 42nd International Conference on Very Large Data Bases.
Paper: http://dx.doi.org/10.14778/2977797.2977800
ArXiv: https://arxiv.org/abs/1605.05219
Poster: https://zenodo.org/record/61653 (doi 10.5281/zenodo.61653)
Gumbo Software: https://github.com/JonnyDaenen/Gumbo
Abstract
While services such as Amazon AWS make computing power abundantly available, adding more computing nodes can incur high costs in, for instance, pay-as-you-go plans while not always significantly improving the net running time (aka wall-clock time) of queries. In this work, we provide algorithms for parallel evaluation of SGF queries in MapReduce that optimize total time, while retaining low net time. Not only can SGF queries specify all semi-join reducers, but also more expressive queries involving disjunction and negation. Since SGF queries can be seen as Boolean combinations of (potentially nested) semi-joins, we introduce a novel multi-semi-join (MSJ) MapReduce operator that enables the evaluation of a set of semi-joins in one job. We use this operator to obtain parallel query plans for SGF queries that outvalue sequential plans w.r.t. net time and provide additional optimizations aimed at minimizing total time without severely affecting net time. Even though the latter optimizations are NP-hard, we present effective greedy algorithms. Our experiments, conducted using our own implementation Gumbo on top of Hadoop, confirm the usefulness of parallel query plans, and the effectiveness and scalability of our optimizations, all with a significant improvement over Pig and Hive.
This first lecture describes what EMT is. Its history of evolution. Main personalities how discovered theories relating to this theory. Applications of EMT . Scalars and vectors and there algebra. Coordinate systems. Field, Coulombs law and electric field intensity.volume charge distribution, electric flux density, gauss's law and divergence
State of ICS and IoT Cyber Threat Landscape Report 2024 previewPrayukth K V
The IoT and OT threat landscape report has been prepared by the Threat Research Team at Sectrio using data from Sectrio, cyber threat intelligence farming facilities spread across over 85 cities around the world. In addition, Sectrio also runs AI-based advanced threat and payload engagement facilities that serve as sinks to attract and engage sophisticated threat actors, and newer malware including new variants and latent threats that are at an earlier stage of development.
The latest edition of the OT/ICS and IoT security Threat Landscape Report 2024 also covers:
State of global ICS asset and network exposure
Sectoral targets and attacks as well as the cost of ransom
Global APT activity, AI usage, actor and tactic profiles, and implications
Rise in volumes of AI-powered cyberattacks
Major cyber events in 2024
Malware and malicious payload trends
Cyberattack types and targets
Vulnerability exploit attempts on CVEs
Attacks on counties – USA
Expansion of bot farms – how, where, and why
In-depth analysis of the cyber threat landscape across North America, South America, Europe, APAC, and the Middle East
Why are attacks on smart factories rising?
Cyber risk predictions
Axis of attacks – Europe
Systemic attacks in the Middle East
Download the full report from here:
https://sectrio.com/resources/ot-threat-landscape-reports/sectrio-releases-ot-ics-and-iot-security-threat-landscape-report-2024/
DevOps and Testing slides at DASA ConnectKari Kakkonen
My and Rik Marselis slides at 30.5.2024 DASA Connect conference. We discuss about what is testing, then what is agile testing and finally what is Testing in DevOps. Finally we had lovely workshop with the participants trying to find out different ways to think about quality and testing in different parts of the DevOps infinity loop.
Epistemic Interaction - tuning interfaces to provide information for AI supportAlan Dix
Paper presented at SYNERGY workshop at AVI 2024, Genoa, Italy. 3rd June 2024
https://alandix.com/academic/papers/synergy2024-epistemic/
As machine learning integrates deeper into human-computer interactions, the concept of epistemic interaction emerges, aiming to refine these interactions to enhance system adaptability. This approach encourages minor, intentional adjustments in user behaviour to enrich the data available for system learning. This paper introduces epistemic interaction within the context of human-system communication, illustrating how deliberate interaction design can improve system understanding and adaptation. Through concrete examples, we demonstrate the potential of epistemic interaction to significantly advance human-computer interaction by leveraging intuitive human communication strategies to inform system design and functionality, offering a novel pathway for enriching user-system engagements.
UiPath Test Automation using UiPath Test Suite series, part 3DianaGray10
Welcome to UiPath Test Automation using UiPath Test Suite series part 3. In this session, we will cover desktop automation along with UI automation.
Topics covered:
UI automation Introduction,
UI automation Sample
Desktop automation flow
Pradeep Chinnala, Senior Consultant Automation Developer @WonderBotz and UiPath MVP
Deepak Rai, Automation Practice Lead, Boundaryless Group and UiPath MVP
LF Energy Webinar: Electrical Grid Modelling and Simulation Through PowSyBl -...DanBrown980551
Do you want to learn how to model and simulate an electrical network from scratch in under an hour?
Then welcome to this PowSyBl workshop, hosted by Rte, the French Transmission System Operator (TSO)!
During the webinar, you will discover the PowSyBl ecosystem as well as handle and study an electrical network through an interactive Python notebook.
PowSyBl is an open source project hosted by LF Energy, which offers a comprehensive set of features for electrical grid modelling and simulation. Among other advanced features, PowSyBl provides:
- A fully editable and extendable library for grid component modelling;
- Visualization tools to display your network;
- Grid simulation tools, such as power flows, security analyses (with or without remedial actions) and sensitivity analyses;
The framework is mostly written in Java, with a Python binding so that Python developers can access PowSyBl functionalities as well.
What you will learn during the webinar:
- For beginners: discover PowSyBl's functionalities through a quick general presentation and the notebook, without needing any expert coding skills;
- For advanced developers: master the skills to efficiently apply PowSyBl functionalities to your real-world scenarios.
Elevating Tactical DDD Patterns Through Object CalisthenicsDorra BARTAGUIZ
After immersing yourself in the blue book and its red counterpart, attending DDD-focused conferences, and applying tactical patterns, you're left with a crucial question: How do I ensure my design is effective? Tactical patterns within Domain-Driven Design (DDD) serve as guiding principles for creating clear and manageable domain models. However, achieving success with these patterns requires additional guidance. Interestingly, we've observed that a set of constraints initially designed for training purposes remarkably aligns with effective pattern implementation, offering a more ‘mechanical’ approach. Let's explore together how Object Calisthenics can elevate the design of your tactical DDD patterns, offering concrete help for those venturing into DDD for the first time!
A tale of scale & speed: How the US Navy is enabling software delivery from l...sonjaschweigert1
Rapid and secure feature delivery is a goal across every application team and every branch of the DoD. The Navy’s DevSecOps platform, Party Barge, has achieved:
- Reduction in onboarding time from 5 weeks to 1 day
- Improved developer experience and productivity through actionable findings and reduction of false positives
- Maintenance of superior security standards and inherent policy enforcement with Authorization to Operate (ATO)
Development teams can ship efficiently and ensure applications are cyber ready for Navy Authorizing Officials (AOs). In this webinar, Sigma Defense and Anchore will give attendees a look behind the scenes and demo secure pipeline automation and security artifacts that speed up application ATO and time to production.
We will cover:
- How to remove silos in DevSecOps
- How to build efficient development pipeline roles and component templates
- How to deliver security artifacts that matter for ATO’s (SBOMs, vulnerability reports, and policy evidence)
- How to streamline operations with automated policy checks on container images
Smart TV Buyer Insights Survey 2024 by 91mobiles.pdf91mobiles
91mobiles recently conducted a Smart TV Buyer Insights Survey in which we asked over 3,000 respondents about the TV they own, aspects they look at on a new TV, and their TV buying preferences.
Welocme to ViralQR, your best QR code generator.ViralQR
Welcome to ViralQR, your best QR code generator available on the market!
At ViralQR, we design static and dynamic QR codes. Our mission is to make business operations easier and customer engagement more powerful through the use of QR technology. Be it a small-scale business or a huge enterprise, our easy-to-use platform provides multiple choices that can be tailored according to your company's branding and marketing strategies.
Our Vision
We are here to make the process of creating QR codes easy and smooth, thus enhancing customer interaction and making business more fluid. We very strongly believe in the ability of QR codes to change the world for businesses in their interaction with customers and are set on making that technology accessible and usable far and wide.
Our Achievements
Ever since its inception, we have successfully served many clients by offering QR codes in their marketing, service delivery, and collection of feedback across various industries. Our platform has been recognized for its ease of use and amazing features, which helped a business to make QR codes.
Our Services
At ViralQR, here is a comprehensive suite of services that caters to your very needs:
Static QR Codes: Create free static QR codes. These QR codes are able to store significant information such as URLs, vCards, plain text, emails and SMS, Wi-Fi credentials, and Bitcoin addresses.
Dynamic QR codes: These also have all the advanced features but are subscription-based. They can directly link to PDF files, images, micro-landing pages, social accounts, review forms, business pages, and applications. In addition, they can be branded with CTAs, frames, patterns, colors, and logos to enhance your branding.
Pricing and Packages
Additionally, there is a 14-day free offer to ViralQR, which is an exceptional opportunity for new users to take a feel of this platform. One can easily subscribe from there and experience the full dynamic of using QR codes. The subscription plans are not only meant for business; they are priced very flexibly so that literally every business could afford to benefit from our service.
Why choose us?
ViralQR will provide services for marketing, advertising, catering, retail, and the like. The QR codes can be posted on fliers, packaging, merchandise, and banners, as well as to substitute for cash and cards in a restaurant or coffee shop. With QR codes integrated into your business, improve customer engagement and streamline operations.
Comprehensive Analytics
Subscribers of ViralQR receive detailed analytics and tracking tools in light of having a view of the core values of QR code performance. Our analytics dashboard shows aggregate views and unique views, as well as detailed information about each impression, including time, device, browser, and estimated location by city and country.
So, thank you for choosing ViralQR; we have an offer of nothing but the best in terms of QR code services to meet business diversity!
GraphRAG is All You need? LLM & Knowledge GraphGuy Korland
Guy Korland, CEO and Co-founder of FalkorDB, will review two articles on the integration of language models with knowledge graphs.
1. Unifying Large Language Models and Knowledge Graphs: A Roadmap.
https://arxiv.org/abs/2306.08302
2. Microsoft Research's GraphRAG paper and a review paper on various uses of knowledge graphs:
https://www.microsoft.com/en-us/research/blog/graphrag-unlocking-llm-discovery-on-narrative-private-data/
Accelerate your Kubernetes clusters with Varnish CachingThijs Feryn
A presentation about the usage and availability of Varnish on Kubernetes. This talk explores the capabilities of Varnish caching and shows how to use the Varnish Helm chart to deploy it to Kubernetes.
This presentation was delivered at K8SUG Singapore. See https://feryn.eu/presentations/accelerate-your-kubernetes-clusters-with-varnish-caching-k8sug-singapore-28-2024 for more details.
Securing your Kubernetes cluster_ a step-by-step guide to success !KatiaHIMEUR1
Today, after several years of existence, an extremely active community and an ultra-dynamic ecosystem, Kubernetes has established itself as the de facto standard in container orchestration. Thanks to a wide range of managed services, it has never been so easy to set up a ready-to-use Kubernetes cluster.
However, this ease of use means that the subject of security in Kubernetes is often left for later, or even neglected. This exposes companies to significant risks.
In this talk, I'll show you step-by-step how to secure your Kubernetes cluster for greater peace of mind and reliability.
Encryption in Microsoft 365 - ExpertsLive Netherlands 2024Albert Hoitingh
In this session I delve into the encryption technology used in Microsoft 365 and Microsoft Purview. Including the concepts of Customer Key and Double Key Encryption.
Encryption in Microsoft 365 - ExpertsLive Netherlands 2024
Chapter 5 Thomas 9 Ed F 2008
1. CrnprpR
Applications
of Integrals
OYERVIEW Many things we want to know can be calculated with integrals: the
areasbetweencurves,the volumes and sudaceareasof solids, the lengthsof curves,
the amount of work it takes to pump liquids fiom belowground, the forces againsr
floodgates,the coordinatesof the points where solid objects will balance.We define
all of these as limits of Riemann sums of continuous functions on closed intervals.
that is, as integrals,and evaluatethese limits with calculus.
There is a pattern to how we define the integrals in applications, a pattern
that, once learned,enablesus to define new integralswhen we need them. We look
at specific applications 0rst, then examine the pattern and show how it leads to
integrals in new situations.
AreasBetweenCurves
This section shows how to lind the areas of regions in the coordinate plane bv
integrating the functions that define the regions, boundaries.
The Basic
Formula a Limitof Riemann
as Sums
Supposewe want to find the area of a region that is bounded above by the curve
-y =.l(r), below by the curve ) = g(.r), and on the left and right by the lines
.r : 4 and r : D (Fig. 5.1). The region might accidentallyhave a shapewhose area
w e c o u l d f i n L w i r h g e o m e t r l b u r i l / a n d g a r e i t r b i l r a r ) o n r i n u o ut,u n c r r o n,r
l . c e
usuaily have to flnd the area with an integral.
To see what the integml should be, we lirst approximate the region with,?
basedon a parlition p : {.r0, ... , r,,} of [ri, D] (Fig. 5.2, on
verticalrectangles rr,
the following page). The area of the tth rectangle(Fig. 5.3, on rhe followrng page;
ls
AA1 : tsjcfi1 x width : Ll (c.t) g(ci)l A4.
We then approximatethe area of the region by adding the areasof the ,?rectanqles:
A* f ^A, -Itr,,.r grr.,rlA,:,.
' The region between y: f(x) and
As ll P | -+ 0 the sums on rhe right approachthe limit
y: g ( x ) a n d t h e l i n e sx : a a n d x = b . _ g(,r)l d,r because
/j t/(r)
365
2. Chapter5: Applicationsof Integrals
356
T - s(ck)
f(t)
Stctll
::.: We approximate regionwith
the
perpendicular the x-axis.
to AAk = areaof kth rectangle,f(c*) - g(cr) = height, lxk = width
rectangles
this integral'
/and g arecontinuous. takethearea theregionto be thevalueof
of
We
That is,
lim )-l/(c,)-gtcrtlAxr - fo rf ,r,-
A- Stxt]dx
nP o-- J
Definition
with /(r) > g(x) throughout b], ihenthe area
g [4,
Ifl ancl arecontinuous
the curves : /(*) andy : g(t) ftom a to b is the
y
of the regionbetween
integral [/ - 8 | liom a to D:
o[
rb
U(x)- g(x)ldx. (1)
A=
J
To apply Eq. (1) we take the following steps.
How to Find the Area Between Two Curves
Graph the cunes and draw a tepresentative rectangle. Thts rcveals
l.
which curve is / (upper curve) and which is g (1owercurye). It also
helps find the limits of integration if you do not aheady know them.
,s'I ,=sinr 2. Find the limits of integration.
I--''-
3. Write a formula for f(x) - g(x). Simplify it if you can.
4. lntegrate [f(x) - gQ, fron a to b Tl1e number you get is the area.
H Find the area between lv = sec2 and y : 3i1;5 from 0 to z/4'
x
EXAMPLE 1
Solution
Step t; We sketch the cutves and a vertical rectangle(Fig. 5.4). The upper curve
is the gmph of /(;r) = secz;r: the lower is the graph of g(;) : sipI'
Sfep 2; The limits of integration are already given: a :0' b = r /4'
,'.]].The regionin Example'lwith a -
SteP3: f (x) - 8(x) = sec2;r sin't
rectangle.
typicalapproximating
3. 5.1 AreasBetweenCurves 367
Step 4:
{r' *- sinx)r/x : ftan + cos
-r x]i/a
lo'o
f'.+l,to+:+
-t
L f,
Curves
That Intersect
When a region is detemined by curves that intersect, the intersection points give
the limits of integration.
- x2
Find the area of the region enclosedby the parabola ! :2
EXAMPLE 2
: -y.
and the line y
Solution
Sfep t; Sketch the curves and a vertical rectangle (Fig. 5.5). Identifying the upper
(-{,,f(-r)
-x. The ,{-coordinates
and the lower curyes, we take /(x) :2- g(r):
xz arLd
pointsare the limits of integration.
of the intersecdon
Step 2; We find the limits of integation by solving | :2 - xz and y : -1 ri-
( 1 ,1 )
multaneously for .r.'
-x [ q r l L t/r1 r ) . u r d . q ( f r
2-xz :
x2-x-2:0 R.rir..
(x+l)(x -2):0 Fxrr.'
(a s(r))
-1,
x: x :2. sot.
The region runs from x : - 1 to.r : 2. The limits of integrationare a : - l, b : 2.
Step 3:
5.5 The regionin Example with a
2
typicalapproximatingrectangle.
Step 4
t2 f -2 -t12
- :
o: rf,.,>e(x)ldx I t z + x - x z t d x = l z r + z
- ,I
l' JJ
J-t I r
/ 4 8 / | 1
(o*r-r,l (-'*l*r/
?qq
-
-'2
At
3 2 l
Technofogy The Intersectionof Two Graphs One of the difficult and some-
times frustrating parts of integmtion applications is finding the limits of inte-
gration. To do this you often have to find the zeroes of a function or the
intenection points of two curues.
g(r) using a graphing utility, you enter
To solve the equation /(x):
yr: f@) and y2: g@)
4. Chapter Applications lntegrals
of
5:
368
you
andusethe grapherroutine to find the Pointsof intersection'Altematively'
- g(T) : 0 with a root finder'Try bothprocedures
lun solve*tJ /(:r)
quation
with
and g(x):3-x'
f (x):lnx
hidden
When points of intersectionarc not clearly revealedor you suspect
further use of calculus
behavior,additional work with the graphing utility or
may be necessary,
ISECT
=.79205996845
curves : Inx and lz:3 - x' usinga built-infundion
y1
a) The intersecting
intersection
to find the -
: x
O) Usinga built-inroot finder to find the zero of f(x) Inx 3 +
with ChangingFormulas
Boundaries
points' we partition
If the formula for a bounding curve changesat one or mole
th,gionintosubregionsthatconespondtotheformulachangesandapplyEq.
(1) to eachsubregion.
Find the areaof the region in the fint quadrantthat is bounded
EXAMPLE 3
- 2'
:
aboveby y : .r/i and below by the t-axis and the line y x
solution
is the graphof
Step t.' The sketch(Fig. 5.6) showsthat the region'supperboundary
<2to g(x) =
fromg(;r):0for0 S r
changes
f(; : JV. Theloweiboundary
-' Z tot Z : t : 4 (thereis agreement r : 2)' We partition the region at r : 2
at
rectanglefor eachsubregion'
into subregiJns ani f and tketh a representative
A
5.6 When the formulafor a boundingcurve
changes, area integralchanges match
to
the
(Example3).
'fe limits of integrationfor regionA area : O andb :,2' The lefchand
step 2:
li#t for region B rs a = 2. To find the right-hand limit, we solve
the equations
5. 5.1 AreasBetween
Curves 369
y : rG andy : x - 2 simultaneously r.'
for
EqratcI (.r)rnd
^-
- z
v^
.9(11
a:(x.-2)2:x.2-4x+4 SqLrafeboth
x 2- 5 x + 4 : o Re$'r'ite.
(.r-1)(r-4):0 FacL(r'.
-_1 -_i
Solve.
Only the value.x=4 satisfies equation
the The valuer:1is an
Ji:x-2.
extraneous introduced squaring. right-hand
root by The limit is b:4.
f (x)- S@): -0:
S t e p3 . ' F o r 05 x = 2 :
- e ( x ) : , 6E - ( x - 2q :
)
Ji - x +2
F o r2 a x a 4 : f(x)
Step 4.' We add the areaof subregions and B to find the total area:
A
f)-f4
=
T o t a fr e a | J i d x I l t J x - x + 2 t d x
a
-o - - - - /
J Jz
areaof B
arca of A
'l2
f) f1 -2 14
fil.+l1*'''-v+2,1,
/2 -
?ef,,_o_ (1,,0,',, .' 8)_
_8
r r , r{zt'''-z++1
?,t, : !.
-, ,
33
lntegrationwith Respect y
to
If a region's bounding curves are describedby functions of y, the approximating
rectanglesare horizontal insteadof vertical and the basic formula has y in place
of r.
t
1
I
L*,,,
use tJteformula
In Eq. (2),/always denotes right-hand
the
- stytldy.
e:
l.' vrrt
(2)
curve and I the left-hand curve, so
fO) - e(y) is nonnegative.
6. Chapter 5: Applicationsof lntegrals
370
Find the area of the region in Example 3 by integrating with
EXAMPLE 4
rcspect to ).
Solution
(c()),
).)
x:y+2 Step t; We sketchthe region and a typical horiz,ontalrectanglebasedon a panition
of an interval ofy values(Fig. 5.7). The region's right-handboundary is the line x :
y f 2, so l (y) : y + 2. The left-hand boundary is the curve ir : y2, so g(y) : y2.
Step 2: The lower limit of integration is l = 0. We find the upper lirnit by solving
I : y *2 and r: )2 simultaneously y; for
5.7 lt takestwo integrations find the l+t:l
to
areaof this regionif we integrate with
2=o
,t-)
respectto x It takes only one if we
with respect y (Example
integrate to 4). ( ) + 1 ) ( ) 2 ): 0 Iurt,,
1'-
_t ) S ,i l .
The upperlimit of integmtion D : 2. (The valuey : -1 givesa point of inter-
is
sectlor
belowthe r-axis.)
Step3:
=)+2 y2=2iy )?
/())-s())
Step 4:
v,rt- :
e:
1,,'' srt,rat 1 2 + y - y ' ) l d y
l,'
f r,2 rt l'
: +; 1l
L2Y 3 ln
8 l0
- o.+ t -
4
3
3
This is the result of Example 3, fbund with less work.
Integrals
with Formulas
from Geometry
Combining
andgeometry.
The fastest
way to find an areamay be to combinecalculus
The Area of the Region in Example 3 Found the Fastesi
EXAMPLE5
way
Find the area of the region in Example 3.
So/ution The area we want is the areabetweenthe curye I : Ji,O = x S 4, and
1:rf the x-axis, minus the area of a triangle with base 2 and height 2 (Fig. 5.8):
2 lr2rr2r
/ x r u = f or t a ,
.2
'.2
2 ,11
J .lo
5.8 The areaof the blue regionis the
:3ttl-o -'z:+'
area underthe parabola = 1& minus
,
y
the areaof the triangle. JJ
-
7. Exercises
5.1 371
Moral of Examples3-5 It is sometimes easierto lind the areabetween two
with respect y instead ,r. Also, it may help to combine
curves integrating
by to of
geometry calculus.
and After sketching region,takea moment determine
the to the
bestwav to oroceed.
Exercises
5.1
Find the areasof the shadedregions in Exercises 1-8.
l. 2.
8. Chapter Applications Integrals
of
5:
372
16. y : x 2 - 2 x and Y=r
area.
In Exercises findthetotalshaded
9-12,
and y = -tt2 14x
t7. ) : 12
10.
9,
] 18. y = l - l x ' ano ]:x-14
(-3,5) 19. y = x 4 - 4 x 2 + 4 and Y=a2
I and y:0
20, y = r n @ = 7 , a>0,
_,_,2
5y = x * 6 (How many intersection points are
21. y = {fl nd
there?)
2x3 - (x212)+4
5x 22,y:lx2-41'and ):
2
I
Find the areasof the regions enclosedby the lines and curves
23-30.
Exercises
-3) : r : 0 , a n d) : 3
x:2y2,
23.
24. x = y2 u1t4 a: y 12
and' 4x - Y = 16
25. y2 - 4x:4
26, x - y2 :0 arld x +2Y2 =3
^nd r+3Y2:2
2 7 ,x * y 2 : O
and ,r*)a:2
2 8 .x - y 2 / t : 0
(-2,4
y
29.x =y2-l and x='yJl
3 0 . r = ) l 3- y 2 a n d x = 2 y
-2 -l Find the areasof the regions enclosedby the curves in Exercises
I 31'34.
and .xa-Y:l
31.4x2+y:4
I a n d 3 x 2- y : 4
y :O
3 2 .x 3
I, (3,5) for xZ0
and :r*)a:1,
3 3 .x * 4 y 2 : 4
and 4.t*y2:0
3 4 ,x a y z : 3
12.
Find the arcas of the regions enclosedby the lines and curves in
)
I+.
35-42.
Exercises
(3,6)
6
and y = sin2.r, 0:.x57t
3 5 .y = 2 s 1 n x
and y:5e92.r, -7t13=x Sr13
v 36.):8cos.r
.t2
and 1:1
3 7 .y : c o s ( r x / 2 )
38. y = sin(nx/2) and 1t:r
(3,1) and'x:tt/4
39, l:sec2.r, y=tarfx, x:-tr/4,
and .x:-tan2), -7r14=r 3n/4
4O. :ta#y
x
r and -x=0, 0=y<7t/2
4 1 .r : 3 s i n l / i o s y
-'
4 2 .y : s s s 2 1 1 t a J 3 1 r r d y : r 1 / 1 , - l = i r : l
3) a
rcgion enclosed the curve
by
43. Find the areaof the propeller-shaped
Find the areasof the regions enclosedby the lines and curves in - y3 : 0 andthe line.r - Y : 0.
t
13-22.
Exercises region enclosedby the
44. Find the area of the propeller-shaped
1 3 . y : a 2- 2 clruesx -yll3 : 0 andr - )l/5 = 0.
ar'd y:2
14. y =2x - x2 arLd Y: -J 45, Find the areaof the region in the first quadrantboundedby the
line I :1, the line,t :2, the curve1 : 1/-r'?, the -t-aris.
and
15. y :1+ and ) :8r
9. Exercises
5.1 373
Suppose area of the region betweenthe graph of a positive
46. Find the area of the tdangular region in the first quadrant the
continuousfunction / and the .x-axisfrom x : 4 to x : b is 4
boundedon the left by the y-axis and on the right by the curves
units. Find the areabetween curves)r: /(.r) and
the
) = sin.rand): cos,r. square
y-2f(x)ftomx:atox-b.
belowby the pambola : ,r2 andabove by
47, The regionbounded )
Which of the following integals, if either calculates areaof
the
: 4 is to be parlitioned into two subsections equal
of
the line )
region shownhere?Give reasonsfor your answer.
areaby cutting acrossit with the horizontal line ), = c. the shaded
a) Sketchthe region and draw a line ) : c acrossit that looks fL f)
-l-x)d:(=
I (x 2xdx
al |
about right. In terms of c, what are the coordinates the
of J
J-t I
points where the line and parabolaintersect?Add them to '
rot tl
-zxax
bt / r-x-(r))dx-
your figure. |
JI
J-t
b) Find c by integratingwith respectto ),. (This puts c in the
limits of integration.)
to.x.(Thisputsc into the
c) Find c by integntingwith respect
integrand well.)
as
''2 and
Find the areaof the regionbetween curve )):3
the
the line y = 1 by integratingwith respectto (a) t, (b) ).
49. Find the areaof the region in the first quadrantboundedon the
left by the ),-axis,below by the line y = t /4, aboveleft by the
curve y = I * .,[, and aboveright by he (i.jtr..te : 2/ Ji,
y
50. Find the area of the region in the first quadrantboundedon the
left by the y-axis, below by the curve r : 2/t, aboveleft by
right by the line i :3 - ).
the curyer : () - l)'?,andabove
54. True, sometimestue, or never tue? The area of the region
between graphsof the continuous functions): /(i) and
the
: g(r) andthe verticallinesr : a andx: b (a < b) rs
)
rb
I lf @) - s(x)ld)c.
J
Give reasonsfor your answer.
S CASExplorationsand Proiects
In Exercises 55-58, you will find the area between curves in the plane
when you cannot find their points of intersectionusing simple algebra.
Use a CAS to perform the following steps:
The figure here shows triangle AOC inscribed in the region cut
a) Plot the curves together to see what they look like and how many
ftom the parabola ) - .r2 by the line y: a2. Find the limit of
(hey hae.
points of inrersection
the latio of the area of the triangle to the area of the parabolic
b) Use the numerical equation solver in your CAS to find all the
region as 4 approaches zerc.
points of intersection.
- g(r)l over consecutive pairs of intersection
c) Integate
f(r)
values.
d) Sum together the integrals found in part (c).
^- ^ -2, I . g 1 x .:1 - J
x
5 5 . '/3 2 3 :
1x1
t4
1 0 .g ( . r r : 8 - l 2 r
5 6 ./ t x t : i - t r ' -
57. /(r) =x *sin(2tc), g(x) : 13
58. /(.x) : -r2sqsir, g(tc) : x1 - x
10. Chapter 5: Applicationsof Integrals
374
Volumes Slicing
by
Finding
From the areasof regions with curved boundaries,we can calculatethe volumes of
cylinders with curved basesby multiplying base area by height. From the volumes
of such cylinders, we can calculate the volumes of other solids.
5licing
Supposewe want to find the volume of a solid like the one shown in Fig. 5.9. At
Cross sectionR(jr). Its area is A(ir).
each point r in the closed inteNal la, rl the cross section of the solid is a region
R(x) whose area is A(r). This makesA a real-valuedfunction of .r' If it is also a
continuous lunction of .r, we can use it to define and calculate the volume of the
solid as an integral in the following way.
We partition the interval [4, ]l along the r-a{is in the usual manner and slice
the solid, as we would a loaf of bread, by planes petpendicular to the -r-axis at
the partition points. The tth slice, the one between the planes at xk r and tn, has
approximatelythe same volume as the cylinder betweenthesetwo planes basedon
the region R(,rr) (Fig. 5.10). The volume of this cylinder is
Vr : base area x height
sedion
lf the areaA(x) of the cross
: A(rr) x (distancebetween the planes at -rr r and xr)
functionof x, we can
R(x)is a continuous
find the volumeof the solid bY : A(xr)Axr.
A(x) from a to b.
integrating
The volume of the solid is therelbre approximatedby the cylinder volume sum
- r,. r,r-.
?-,'^'^
This is a Riemann sum for the function A(r) on [c, Dl. We expectthe approxlmahons
from these sums to improve as the norm of the partition of la, bl goes to zero, so
we define their limiting integral to be the volume of the solid.
Approxrmatlng
Plane at.r[ I cllinderbased p l a n en r r ,
on]((rt]
-'i.---------'
-,
The cylinder's base
is the region R(,v*).
NOTTO SCAI-E
view of the sliceof the solidbetweenthe
. Enlarged
cylinder'
planes xk r and xk and its approximating
at
11. 5 . 2 F i n d i n g o l u m eb y S l i c i n g 3 7 5
V s
Definition
The volume of a solid of known integrable cross-sectionarea A(x) from
x = a to x: D is the integralofA from a to r.
v : fo e61a,. (j)
J
T o a p p l l E q . { l ) . w e r a k et h e f o l l o w i n g: r e p . .
How to FindVolumesby the Method of Slicing
1 Sketch the solid and a typical cross section.
2. Find a formula for A(r).
3. Find the limits of integration.
4. Integrate A(x) to find the volume.
EXAMPLE 7 A pyramid 3 m high has a square base that is 3 mon a side.
The cross section of the pyramid perpendicularto the altitude r m down from the
ve ex is a squarer m on a side. Find the volume of the pyramid.
Solution
Step 1: A sketcll. We draw the pyramid with its altitude along the r-axis and its
vertexat the origin and includea typical crosssection(Fig. 5.11).
Typical cfoss
section
--r$€t
-=_,,.,
sections the pyramidin Example
The cross of 1
are 5quare5.
Step 2: A.fonnula for A(x). The cross section at r is a squarex meters on a side,
so lts area ls
A(r) : 12'
Step 3: Tlle lintits of integration. The squaresgo from r :0 to;r :3
Step 4: The volwne.
'l'
lb t' - r.
-,
At^trlt -
v=I | t:dr r |=9
J.. J,, .l ,
The volume is 9 mr.
12. Chapter Applications Integrals
of
376 5:
wedge cut from a cylinderof radius3 by two planes.
is
2 A curved
EXAMPLE
Bonaventura Cavalieri ('l 598-1647) planecrosses
One planeis perpendicular the axis of the cylinder.The second
to
Cavalied, a studentof Galileo's, discovered the nrst plane at a 45' angle at the centerof the cylinder. Find the volume of the
that if two plane rcgions can be afiangedto wedge.
lie over the sameinterval of the i-axis 1n
sucha way that they haveidentical vertical Solution
crosssectionsat every point, then the rcgions perpen-
Step 1: A sketch. draw the wedgeand sketcha typical crosssection
We
havethe sarnearea.The theorem(and a letter
dicularto the:r-axis(Fig. 5.12).
from Galileo) were
of recommendation
enoughto win Cavalied a chai at the
Univelsityof Bolognain 1629.The solid
2lt -7
geometryversionin Example 3, which
Cavalieri neverproved,was given his name
by later geometen.
have
Crcsssections
the samelongthat
everypoint in [d, b]
5-t2 The wedgeof Example slicedperpendicular
2.
sections redangles.
are
to the x-axis.
The cross
Step 2: Theformula for A(x). The cross section at r is a rectangleof area
A(') : (heishtxwidth) (zJe - 'r)
: (r)
: z x t t: Y - x ' .
^,
Step 3: The linxits of integration. The rectangles from r :0 to.r :3.
run
Step 4: The volune.
?b ^3
Ax)dx: | 2tcJ9x2dx
-
v:l
Ja JO
113
: -;e - xr3/'
)o
I
,f : o+?e),/' 1! = lf r1r. nncsfalc-
J . L n d s u b s t i l U r eb r c k .
tr
- 19
Cavalieri's Theorerrr
EXAMPLE 3
Cavalieri's theoremsaysthat solids with equal altitudesand identical parallel cross-
theorem. Thesesolids
5.13 Cavalieri's
section areashave the samevolume (Fig. 5.13). We can seethis imrnediately from
havethe samevolume.You can illustrate
Eq. (1) becauset}Ie cross-sectionarea function A(.r) is the same in each case. f
this yourself
with stacks coins.
of
13. Exercises
5.2 377
Exercises
5.2
Areas
Cross-Section
In Exercises and 2, find a fomula for the arcaA(r) of the cross
I
sectionsof the solid perpendicular the l-axrs.
to
1. The solid lies betweenplanesperpendicular the r-axis at i =
to
-l and,r:1. In eachcase,the crosssections perpendicular
to the r-axis betweentheseplanesrun from the semicircley =
-^4 - ,r2 to the semicircley : 4[ - ,2.
a) The cross sectionsme circular disks with diametersin the
xy-plane.
planesperpendicular ther-axis at.r : 0
The solidliesbetween to
andl : 4. The crosssectionsperpendicular the-t-axisbetween
to
these planes run from the parabola y = -aE to the parabola
r: Ji.
a) The cross sectionsare circular disks with diametersin the
r)-plane.
b) The crcsssections squares
are with bases ther),-plane.
in
b) The crosssectionsare squareswith basesin the r)-plane.
c) The cross sectionsarc squarcswith diagonalsin the xy-
plane. (The length of a square'sdiagonalis .,4 times the
lengthof its sides.)
c) The crosssections squares
are with diagonals ther)-plane.
in
d) The crosssectionsare equilateraltriangleswith basesin the
rJ-plane.
Volumes Slicing
by
Find the volumesof the solids in Exercises3-12.
3. The solid lies betweenplanesperpendicular the,-axis at .r : 0
to
and r :4. The crosssections perpendicular the axis on the
to
The crosssectionsarc equilateraltriangleswith bases the
in interval 0 5 i :4 are squareswhose diagonalsrun ftom the
d)
parabola1 : -.vE to the parabolaf : Jtr.
ry-plane.
14. 374 chapter5: Applications Integrals
of
4, The solid lies between planes perpendicularto the r-axis at -x :
- 1 and r : I . The cross sections perpendicular to the ir axis are
run from lhe parabolay - x7 to
circulardisks whosediamerers
the Darabola :2 - t2.
t
5. The solid lies betweenplanesperpendicular the r-axis at
to Cavalieri's
Theorem
r : - I andr : 1. The crosssections
perpendicular the axis
to
ll. A twisted solid. A square sidelengths lies in a planeper-
of
betweentheseplanesare vertical squares whosebase edges
pendicular a line L Onevertex the square on t. As this
to of lies
run from the semicircle - - 1(1- rt lo rhe semicircle -
y
J
squ&emovesa distance alongl, the square
l? tums one revo-
!7-7.
lution aboutZ to generatea corkscrewlike column with square
perpendicular ther-axis at r =
6. The solidlies between planes to crosssections.
I and n: l. The crosssections perpendicular the ,r-axis
to
a) Find the volume of the column.
planes squares
between these are whosediagonals frcm the
run
b) What will the volumebe if the squaretums twice instead
semicircle : -{ -7 to the semicircle
y: JT=V. Qn
) Givercasons your answer
of once? for
lengthof a square's
diagonal r/Z timesthelengthof its sides.)
is
berween curve) : 2 vfi; planes
12. A solid lies between
7. t he base thesolidis thereSion
of lhe
The qoss sections perpendicular tbe -{-axisat
and the interval[0, r] on the.r-axis. perpen- to
-r = 0 ard .r : 12.The cross
dicularto the r-axis are
sections planes perpendicular
by
to ther-axis arecirculardisks
whosediametersrun from the line
) : -r/2 to the line ) : -r.
Explain why the solid has the
same volumeas a right circular
conewith baseradius3 and
height12.
13. Cavalieri'soriginal theorem. ProveCavalieri's
original theo
veftical equilateral triangles with bases running from the
a)
rem (marginalnote, page 376), assuming
that each region is
I-axrs to me culve:
boundedabove belowby thegraphs continuous
and of functions.
b) vertical squareswith basesrunning from the r-axis to the
t4. The volume of a hemisphere(a classical
cutve. application af Cav-
alieri's theorem). Derive the formula V : (2/3)zR3 for the
8. The solid lies between planes pe.pendicularto the r-axis at -x =
volumeof a hemisphere radiusR by comparing crosssec-
of its
-r /3 and x : xr/3. The crcss sectionsperpendicularto the
r-
tionswith the crosssections a solid right circularcylinderof
of
axis are
mdiusR andheightR from which a solidright circularconeof
circular disks with diameters running from the curve y :
a)
baseradius andheishtR hasbeenremoved.
R
tan,r to the curve ): secr;
verlical squares whose base edges run from the curve y :
b)
tanr to the cuNe ): SeCr.
9. The solid lies betweenplanesperpendicularto the y-axis at y : 0
and ) : 2. The cross sectionsperpendicularto the ]-axis are cir-
cular disks with diametersrunning from they-axis to the parabola
10. The base of the solid is the disk j'? + ),2 5 1. The cross sections
I and y - I
by planesperpendicularto the )-axis between], -
are isoscelesdght triangles with one leg in the disk.
15. 5.3 Volumesof Solidsof Revolution-Disksand Washers 379
rn::rt sotids Revorution-Disks
of
E n
The most common applicarion of the method of slicing is to solids of revolution.
Solids of revolution are solids whose shapescan be generatedby revoJvingplane
regions about axes. Thread spools are solids of revolution; so are hand weights
and billiard balls. Solids of revolutign sometimeshave volumes we can find with
formulas liom geometry, as in the case of a billiard ball. But when we want to
find the volume of a blimp or to predict the weight of a part we are going to have
turned on a lathe, formulas from geometry are of little help and we tum to calculus
lor lhe un:* ers.
6
revolution
Axisof
revolution
s
#
If we can arange for the region to be the region betweenthe graph of a contin-
uous function I = R(r), a 5 x < ,, and the x-axis, and for the axis of revolution
to be the x-axis (Fig. 5.14), we can find the solid's volume in the following way.
The typical cross section of the solid perpendicularto the axis of revolution is
a disk of radius R(x) and area
A(.r) : 71136iut1: rlR(x)12.
:
The solid's volume, being the integral ofA from.r - a to x : b, is the integral of
Cross sectionperpendjcularto
zfR(x)1'zfrom a to D.
the axis at r is a disk ofarea
,4(r) = 7r(radius)2 tt (R(,r))2
=
Volume of a Solid of Revolution (Rotation About the .r-axis)
The volume of the solid generatedby revolving about the.{-axis the region
betweenthe -r-axis and the graph of the continuousfunction ) : R(r), a S
:r 5 b, is
v = [ , ftua;ur1t : [' r l R 1 x ) l 2 l r .
a* (1)
z
J. J,,
EXAMPLE I The region betweenthe curye y :.,rE,O <.{ S 4, and the r-axis
and the x-axisfrom a to b about the
x-axis. is revolved about the .x axis to generatea solid. Find its volume.
16. 380 Chapter Applications lntegrals
5: of
:l,
Solution We draw ligures showing the region, a typical radius, and the generated
solid (Fig. 5.15).The volumeis
fb
v : I trlR(x)'dx F - qr l r
f1
: I rlrtTax /l1ft i
Jo-
(4t2
f4
- u . ,t 2 l a
=tr
lo x d x LJo =z-:8n.
1
l
J
,, .f
How to Find Volumes Using Eq. (1)
1, Draw the regionandidentifythe radiusfunctionR(x).
2. Square R(r) andmultiply by z.
3. Intesrate find the volume.
to
The axis of revolution the next example not the .x-axis, the rule for
in is but
calculating volumeis thesame:
the Integmtelr(radius)2
betweenappropriate
limits.
(b)
5.15 The region(a) and solid(b) in EXAMPLE2 Find the volumeof the solid generated revolvingthe region
by
Exampl1.e
bounded y : .rf andthe lines) : l, x:4 about line y : 1.
by the
Solution We drawfiguresshowing region,a typicalradius,andthe generated
the
solid (Fig. 5.16).The volumeis
F
f( lrr
V: I r l 'R ( x ) 1 d r
z
J,
r1
: I n l' J x _ t ) - d x
Jt
f4
: o I 'l , z E+ r ) a ,
Jt
^ .4
^ tt tn
lx- - + x l : -I
:Tl 2':x
^
J o
Lz lr -j
To find the volume of a solid generated by revolving a region between the
)-axis and a curve x : R()), c < l : d, about the y-axis, we use Eq. (i) with r
replaced by ).
Volume of a Solid of Revolution (Rotation About the y-axis)
(b)
5.16 The region(a) and solid(b) in
Example2.
-
17. 5.3 Volumesof Solidsof Revolution_Disksand Washers 391
EXAMPLE 3 Find the volume of the solid generatedby revolving Lnereglon
betweenthe y-axis and the curve x:2/y,I S l,:4, aboutthe y-axis.
Solution We draw figures showing the region, a typical radius, and the generated
solid (Fig. 5.17).The volumeis
v: :zln6)1'z
ay l_ l (lr
r
l,o
= ;) o,
t4 /'r2
J, /11 I
r
r^ 4 T r'1.
l3l
I - tlt : qr | -: | : 4n r . t
-t
Jt v L )lr L4J
.l
EXAMPLE 4 Find the volume of the solid generatedby revolving the region
betweenthe parabola :12 + I and the line I :3 aboutthe line; :3.
I
so/ut on We draw ligures showing the region, a typical radius, and the generated
solid (Fig. 5.18). The volume is
( N1
,: l.r r: )
J nrlR(y)'?dy
pD
/11r:1 rf -ll
rl2 - y212
: dy
I
(b)
/1
:r
5.17 The region(a) and solid(b) in L4-4y2+y4ldy
I
Example3.
.,,51f
: 7 r | 4 y - ; - y '+ a I
4
l J )l^
I
64rJ2
t5
R(])=3-()'?+ 1)
)
z
0
-.'t
r:t-+r
I
(a.) (b)
5.78 The region(a) and solid(b) in
Example4.
l
18. Chapter 5: Applicationsof lntegrals
382
The WasherMethod
If the region we revolve to generate a solid does not border on or cross the axis of
revolution, the solid has a hole in it (Fig. 5.19). The cross sectionsperpendicular
to the axis of revolution ate washersinstead of disks. The dimensionsof a typical
washer are
R(-r)
Outerradius:
, rx )
Inner radius:
areais
The washer's
A(x) : TtlR(;)lz rVG)12: o (tn{)l' - tt(r)lt).
-
The Washer Fonnula for Finding Volumes
fb
.lrtx)f)dx
v: I (3)
ra ( [ R r x ) | ' : |
|
tl
outer inner
radius
mdius
squarcd squarcd
) = R(-r)
Notice that the function integrated in Eq. (3) is t(Rz - r2), not T (R - r)'z. Also
notice that Eq. (3) gives the disk method fomula if /(jr) is zero throughout [a, r].
Thus. the disk method is a soecial case of the washer method.
The region boundedby the curve y : xz + | and the line y :
EXAMPLE 5
-x t 3 is revolved about the -r-axis to generatea solid. Find the volume of the
solid.
5.?9 The crosssectionsof the solid of
revolution generated here are washers, Solution
not disks, the integralf Ak) dx leads
so
Sfep t; Draw the region and sketcha line segmentacrossit perpendicularto the
to a slightlydifferentformula.
(the red segment Fig. 5.20).
in
a;dsof revolution
Step 2,' Find the limits of integrationby finding the.r-coordinates the intersection
of
points.
x2+l:-x+3
,'+*-2:o
(x+2)(x-l):0
, -_1 .-l
Step 3; Find the outer and inner radii of the washerthat would be swept out by
the line segmentif it were revolved about the x-axis along with the region. ftVe
drewthe washer Fig. 5.21,but in your own work you neednot do that.)These
in
radii are the distances the endsof the line sesmentfrom the axis of revolution.
of
R(x) : -1 -1-
3
Outer radius:
r(x):a241
Inner radius:
19. 5.3 Volumes Solids Revolution-Disks Washers 383
of of and
+-1
(1,2)
lntegratlon
j . . i T h er e g i o ni n E x a m p l 5 s p a n n e d y
e b
a line segmentperpendicular the axis to
of revolution. when the regionis 5.21 The inner and outer radii of the
cmss section
Washer
about the x-axis, line
revolved the Outerradius: = r+3 washersweptout by the line segmentin
R(r)
will generatea washer.
segment r(.r) = 12 + I
Innerradius: Fi9.5.20.
Step 4; Evaluate the volume integral.
rb
-
,= ([Rrx)l' lr(x)]')
dx F ( t 1 - il
J,,
lt ^
= ({ x | 3)' tx' I lt'tdt
I ltn!1 l
/_rtr
fl
= / z(S 6x x2 xa;dx
J-2
r' r'l' tllr
-'l*t^
i sl, s LI
How to FindVolumesby the WasherMethod
Draw the region and sketch a line segmentacross it perpend.icularto
l.
the q,is of revolution. When the region is revolved, this segmentwill
generatea typical washer cross section of the generatedsolid.
2. Find the limils of integration.
Find the outer qnd inner rqdii of the washer swept out by the line
segment.
4. Integrate to flnd the volume.
To find the volume of a solid generatedby revolving a region about the )-axis,
we use the stepslisted above but integratewith respectto l instead of ,r.
The regionboundedby the parabola : x2 and the line y : 2y
EXAMPLE 6 I
quadrantis revolved about the )-axis to generatea solid. Find the volume
in the first
of the solid.
20. 384 Chapter Applications Integrals
5: of
Solution
Step t.' Draw the region and sketch a line segment across it perpendicular to the
axis of revolution, in this case the ),-axis (Fig. 5.22).
Step 2: The line and parabola intersect at ) : 0 and l : 4, so the limits of inte-
grationare c :0 andd :4.
Step 3.' The radii of the washer swept out by the line segment are R(y) : a/ry,
r(y) : y/2 (Figs.5.22 and 5.23).
?
t22 The region,limitsof integration, 523 The washersweptout by the line
and radii in Example
6. segmentin Fig.5.22.
Step 4:
Eq. (-l ) 'irh i
-
,: , (tnci)l, tr(y)t2)
dy
1
: l,^ -ltrJ')r,
(ltl' sleps I .rnd -l
:L'(,-';)*:l+-i]:, 8
3
EXAMPLE 7 The region in the first quadrantenclosedby the parabola y :2e2,
the y-axis, and the line ) : I is revolved about the line,r : 3/2 to generatea solid.
)=*2or*:f
t
5ol
Find the volume of the solid.
4!.=;-ti
Solution
Step 1.' Draw the region and sketch a line segment across it perpendicular to the
Er
axis of revolution, in this case the line ; :3/2 (Fig. 5.24).
Step 2: Tbe limits of integration are ) : 0 to l : 1.
Step 3.' The radii of the washer swept out by the line segment arc R(y) :3/2,
524 The region,limitsof integration,
r(y) : (3/2) - -/1, (Fiss. 5.24 and 5.25).
and radii in Example
7.