1. INTEGRAL CALCULUS
DIFFERENTIATION UNDER THE INTEGRAL SIGN:
Consider an integral involving one parameter ‫ ן‬and denote it as
௕
‫ܫ‬ሺ‫ן‬ሻ ൌ න ݂ሺ‫ן ,ݔ‬ሻ݀‫ݔ‬
௔
where a and b may be constants or functions of ‫ .ן‬To find the derivative of
‫ܫ‬ሺ‫ן‬ሻ when it exists it is not possible to first evaluate this integral and then to
find the derivative, such problems are solved by using the following rules.
(A) Leibnitz’s Rule for Constant limits of Integration:
Let ݂ሺ‫ן ,ݔ‬ሻ ܽ݊݀ ሺ‫ן ,ݔ‬ሻ be continuous functions of x and ‫ ן‬then
డ௙
డ‫ן‬
‫݂ ׬‬ሺ‫ן ,ݔ‬ሻ ݀‫ݔ‬ ൌ‫׬‬ ሺ‫ן ,ݔ‬ሻ݀‫,ݔ‬
ௗ ௕ ௕ డ௙
ௗ‫ ן‬௔ ௔ డ‫ן‬
where a, b are constants and independent of parameter ‫ן‬
If in the integral ‫ ܫ‬ሺ‫ן‬ሻ ‫׬‬ ݂ሺ‫ן ,ݔ‬ሻ ݀‫ݔ‬
(B) Leibnitz’s Rule for Variable Limits of Integration:
௕ሺ‫ן‬ሻ
௔ሺ‫ן‬ሻ
݂ሺ‫ן ,ݔ‬ሻ satisfies the same conditions, and are functions of the
݂ሺ‫ן ,ݔ‬ሻ ݀‫ݔ‬ ൌ‫׬‬ ݂ሺ‫ן ,ݔ‬ሻ ݀‫ ݔ‬൅ ݂ሺܾ, ‫ן‬ሻ െ ݂ሺܽ, ‫ן‬ሻ
parameter‫ ,ן‬then
‫׬‬
ௗ ௕ሺ‫ן‬ሻ ௕ሺ‫ן‬ሻ డ ௗ௕ ௗ௔
ௗ‫ ן‬௔ሺ‫ן‬ሻ ௔ሺ‫ן‬ሻ డ‫ן‬ ௗ‫ן‬ ௗ‫ן‬
***********************************************************
Example 1: Evaluate ‫׬‬ ݀‫ ݔ‬and hence show that ‫׬‬ ݀‫ ݔ‬ൌ
**
ஶ ௘ షೌೣ ୱ୧୬ ௫ ஶ ୱ୧୬ ௫
଴ ௫ ଴ ௫
గ
ଶ
by using differentiation under integral sign.
Solution: Let ‫ܫ‬ሺߙሻ ൌ ‫׬‬ ݀‫ݔ‬
ஶ ௘ షೌೣ ୱ୧୬ ௫
଴ ௫
Differentiate w.r.t ‫ ן‬by Leibnitz’s Rule under integral sign.
݀ ஶ
߲ ݁ ି௔௫ sin ‫ݔ‬
‫׵‬ ൫‫ ܫ‬ሺߙሻ൯ ൌ න ݀‫ݔ‬
݀ߙ ଴ ߲ߙ ‫ݔ‬
ஶ
ൌ න ݁ ି௔௫ sin ‫ݔ݀ ݔ‬
଴
ஶ
ൌ െ ቂ݁ ି௔௫ ሺ ሻቃ then
ିఈ ୱ୧୬ ௫ି௖௢௦௫
ఈమ ାଵ ଴
‫ܫ‬ሺߙሻ ൌ െ‫ ߙ ݊ܽݐ‬൅ ݇ … … … … … . . ሺ1ሻ
ିଵ
ܶ‫ ߙ ݐݑ݌ , ݇ ݐ݊ܽݐݏ݊݋ܿ ݄݁ݐ ݂݀݊݅ ݋‬ൌ ∞ ݅݊ ሺ݅ሻ

2. ߨ ߨ
‫ܫ ׵‬ሺ∞ሻ ൌ െ‫ି݊ܽݐ‬ଵ ∞ ൅ ݇ ൌ െ ൅ ݇ ‫݇׵‬ൌ
2 2
‫ܫ ׵‬ሺߙሻ ൌ െ ‫ି݊ܽݐ‬ଵ ߙ ‫ ߙ ݐݑ݌‬ൌ 0 ‫׬ ݐ݁݃ ݁ݓ‬ ݀‫ ݔ‬ൌ
గ ஶ ୱ୧୬ ௫ గ
ଶ ଴ ௫ ଶ
*****************************************************
݀‫ ߙ , ݔ‬൒ 0
Example 2: using differentiation under the integral sign, evaluate
‫׬‬
ଵ ௫ ೌ ିଵ
଴ ୪୭୥ ௫
Solution: Let ‫ܫ‬ሺߙሻ ൌ ‫׬‬ ݀‫ . . … … … … ݔ‬ሺ1ሻ
ଵ ௫ ೌ ିଵ
଴ ୪୭୥ ௫
‫݄݊݁ݐ‬ ൫‫ ܫ‬ሺߙሻ൯ ൌ ‫ ׬‬డ‫ ן‬୪୭୥ ௫ ݀‫ ݔ‬ൌ ‫׬‬ ݀‫ ݔ‬ൌ ‫ ݔ ׬‬௔ ݀‫ ݔ‬ൌ
ௗ ଵ డ ௫ ೌ ିଵ ଵ ௫ ೌ ୪୭୥ ௫ ଵ
ௗఈ ଴ ଴ ୪୭୥ ௫ ଴
ଵ
ଵାఈ
Integrating both sides w.r.t ߙ
‫ܫ ׵‬ሺߙሻ ൌ logሺ1 ൅ ߙሻ ൅ ‫ . . … … … … … ܥ‬ሺ2ሻ
From ሺ1ሻ when ߙ ൌ 0 ‫ܫ‬ሺ0ሻ ൌ 0
From ሺ2ሻ when Iሺ0ሻൌlog ሺ1ሻ൅C ‫ ܥ ׵‬ൌ 0
The solution is ࡵሺࢻሻ ൌ ‫܏ܗܔ‬ሺ૚ ൅ ࢻሻ
*****************************************************
Example 3: Evaluate ‫׬‬ ݀‫ ݔ‬using the method of differentiation
ஶ ௧௔௡షభ ௔௫
଴ ௫ሺଵା௫ మ ሻ
݀‫ݔ‬
under integral sign.
Solution: Let I = ‫׬‬
ஶ ௧௔௡ షభ ௔௫
଴ ௫ሺଵା௫ మ ሻ
Differentiate w.r.t a by Leibnitz’s rule under integral sign
ஶ
݀ ߲ 1
‫׵‬ ൫‫ ܫ‬ሺܽሻ൯ ൌ න ሺ‫ି݊ܽݐ‬ଵ ܽ‫ݔ‬ሻ ݀‫ݔ‬
݀ܽ ߲ܽ ‫ݔ‬ሺ1 ൅ ‫ ݔ‬ଶ ሻ
଴
ஶ
‫ݔ‬ 1
ൌන . ݀‫ݔ‬
ሺ1 ൅ ܽଶ ‫ ݔ‬ଶ ሻ ‫ݔ‬ሺ1 ൅ ‫ ݔ‬ଶ ሻ
଴
Let ൌ ሺଵା௔మ ൅ ሺଵା௫ మ ሻ by partial fractions
ଵ ஺௫ା஻ ஼௫ା஽
ሺଵା௔మ ௫ మ ሻሺଵା௫ మ ሻ ௫ మሻ
Solving, ‫ ܣ‬ൌ 0, ‫ ܤ‬ൌ , ‫ ܥ‬ൌ 0, ‫ ܦ‬ൌ
௔మ ିଵ
௔మ ିଵ ௔మ ିଵ
ஶ
݀ 1 ܽଶ 1 ߨ
‫׵‬ ൫‫ ܫ‬ሺܽሻ൯ ൌ ଶ නቊ െ ଶሻ ቋ
݀‫ ݔ‬ൌ
݀ܽ ܽ െ1 ሺ1 ൅ ܽ ଶ‫ ݔ‬ଶሻ ሺ1 ൅ ‫ݔ‬ 2ሺܽ ൅ 1ሻ
଴
Integrating w.r.t ܽ ‫ܫ ׵‬ሺܽሻ ൌ ‫ ׬‬ቀ ቁ ݀ܽ ൌ logሺܽ ൅ 1ሻ ൅ C
గ ଵ గ
ଶ ௔ାଵ ଶ
Then C ൌ0 by putting ܽൌ0

3. Example 4: Evaluate ‫݃݋݈ ׬‬ሺ1 ൅ ߙ ܿ‫ݔݏ݋‬ሻ݀‫ ݔ‬using the method of
గ
*****************************************************
଴
differentiation under integral sign
****************************************************************
formulae
ormulae:
Reduction formulae:
Iሻ ‫ ׬‬Sin୬ θ d θ
IIሻ ‫ ׬‬cos ୬ θ d ߠ
IIIሻ ‫ ׬‬sin୫ θcos ୬ θ ߠ
And to evaluate
Iሻ
ಘ
‫ ׬‬మ sin୬ θ d ߠ
଴
IIሻ
ಘ
‫ ׬‬cos ୬ θ d ߠ
଴
మ
IIIሻ ‫׬‬ sin୫ θcos ୬ θ d ߠ
஠/ଶ
଴
∫ sin x dx = ∫ sin n −1 x. sin x dx With ߠ = x
n
(a )
= sin n −1 x.(− cos x) − ∫ ( n − 1) sin n− 2 x. cos x( − cos x) dx
= − sin n −1 x. cos x + ( n − 1) ∫ sin n − 2 x.(1 − sin 2 x)dx
= − sin n−1 x. cos x + ( n − 1) ∫ sin n − 2 xdx − ( n − 1) ∫ sin n x dx
n ∫ sin n x dx = − sin n −1 x. cos x + (n − 1) ∫ sin n− 2 x dx
Or sin n −1 x. cos x n − 1
∫ sin x dx = − ∫ sin x dx ........... (1)
n n−2
+
n n
Similarly, (b) sin x. cos n−1 x n − 1
∫ cos x dx = − n ∫
cos n − 2 x dx ........... ( 2)
n
+
n
Thus (1) and (2) are the required reduction formulae

4. TO Evaluate
Then etc
i) =
ii) ( put x = a sinθ, so that dx = a cosθ dθ
θ
Also when x = 0, θ = 0, when x = a, θ = π/2)
=
= =

5. Evaluate ‫׬‬૙
ஶ ࢊ࢞
࢔
൫ࢇ૛ ା ࢞૛൯
iii) ( put x = a tan θ, so that dx = a sec2θ dθ
Also when x = 0, θ = 0, when x = ∞, θ =
π/2)
‫׬‬૙
࣊/૛ ࢇ ࢙ࢋࢉ૛ ࣂ
ࢇ૛࢔ ࢙ࢋࢉ૛࢔ ࣂ
=
ಘ
‫ ׬‬మ cos ଶ୬ିଶ θ dθ
૚
ࢇ૛࢔ష૚ ଴
=
. ሺ૛࢔ି૛ሻሺࢇ࢔ି૝ሻ………૝.૛ . .
૚ ሺ૛࢔ି૜ሻሺ૛࢔ି૞ሻ………૜.૚ ࣊
ࢇ૛࢔ష૚ ૛
=
Evaluate ‫ ׬‬sinସ x cos ଶ x dx
‫ ׬‬sinଶ x ሺsinx cosxሻଶ dx
iv)
=
ଵିୡ୭ୱଶ୶ ୱ୧୬ଶ୶ ଶ
‫׬‬ ቀ ቁ dx
ଶ ଶ
=
‫׬‬ሺsinଶ 2x െ sinଶ 2x cos2xሻ dx
ଵ
଼
=
ቂ‫׬‬ dx െ ‫ ׬‬sinଶ 2x ሺ cos2x. 2ሻdxቃ
=
ଵ ଵିୡ୭ୱସ୶ ଵ
଼ ଶ ଶ
ቂ‫ ׬‬dx െ ‫ ׬‬cos 4x dx െ dxቃ
ଵ ୱ୧୬య ଶ୶
ଵ଺ ଷ
=
ቂ‫ ܠ‬െ ‫ܖܑܛ‬૝‫ ܠ‬െ ‫ܖܑܛ‬૜ ૛‫ܠ‬ቃ
૚ ૚ ૚
૚૟ ૝ ૜
=
Evaluate ‫׬‬଴ cos ସ θ sinଷ 6θ d ߠ
஠/଺
v)
‫׬‬଴ cos ସ θ ሺ2sin3θcos3θሻଷ d ߠ
஠/଺
=
ૡ ‫׬‬଴ sinଷ θ cos ଻ 3θ d ߠ
஠/଺
=
‫ ׬‬sinଷ x cos ଻ x d x
ૡ ஠/ଶ
૜ ଴
=
. ൌ
ૡ ૛ൈ૟.૝.૛ ૚
૜ ૚૙.ૡ.૟.૝.૛ ૚૞
=
(Put 3ࣂ = x ; so that 3dࣂ = dx. Also when ࣂ = 0, x = 0
When ࣂ = ࣊/6, x = ࣊/2
Evaluate ‫ ݔ ׬‬ସ ሺ1 െ ‫ ݔ‬ଶሻ ݀‫ݔ‬
ଵ ଷ/ଶ
଴
vi)
‫׬‬଴ ‫݊݅ݏ‬ସ ‫ݐ‬ሺܿ‫ ݏ݋‬ଶ ሻଷ/ଶ ܿ‫= ݐ݀ ݐ ݏ݋‬
గ/ଶ
=
‫׬‬଴ ‫݊݅ݏ‬ସ ‫ ݏ݋ܿ ݐ‬ସ ‫ݐ݀ ݐ‬
గ/ଶ
=

6. . . ൌ
૜ ૚ ൈ ૜.૚ ࣊ ૜૟࣊
ૡ ૟. ૝. ૛ ૛ ૛૞૟
=
(put x = sint so that dx = cost dt, when x = 0, t = 0 when x = 1, t = ࣊/2 )
************************************************************************
Tracing of Curves:
For the evaluation of Mathematical quantities such as Area, Length,
Volume and Surface area we need the rough graph of the equation in either
Cartesian or parametric or polar form depending on the statement of the
problem. We use the following theoretical steps to draw the rough graph.
A) Cartesian Curves: y = f (x)
a) Symmetry:
i) If the power of y in the equation is even, the curve is symmetric
about x- axis
ii) If the power of x in the equation is even, the curve is symmetric
about y- axis
iii) If both the powers x and y are even then the curve is symmetric
about both the axis.
iv) If the interchange of x and y leaves the equation unaltered then the
curve is symmetric about the line y = x
v) Replacing x by – x and y by – y leaves the equation unchanged the
curve has a symmetry in opposite quadrants.
b) Curve through the origin:
The curve passes through the origin, if the equation does not contain
constant term.
c) Find the origin, is on the curve. If it is, find the tangents at 0, by equating
the lowest degree terms to zero.
i) Find the points of intersections with the coordinate axes and the tangents
at these points.
.
For, put x = 0 find y; and put y = 0,
ௗ௬
ௗ௫
find x. At these points, find
‫݂ܫ‬ ൌ ∞, then the tangent is parallel to y axis.
ௗ௬
ௗ௫
ൌ 0, then the tangent is parallel to x axis.
ௗ௬
ௗ௫
If
d) Asymptotes: express the equation of the curve in the form y = f (x). Equate
the denominator to zero. If the denominator contains x, then there is an
asymptote.

7. e) Find the region in which the curve lies.
f) Find the interval in which the curve is increasing or decreasing.
B) Parametric Form: x=f(t), y=g(t)
In this case we try to convert the parametric form into Cartesian form by
eliminating the parameter (if possible). Otherwise we observe the following
I) Find dy/dt and dx/dt and hence dy/dx.
II) Assign a few values for t and find the corresponding value for x, y ,y’.
III) Mark the corresponding points, observing the slope at these points.
C) Polar curves: r = f(ࣂ)
a) Symmetry: 1. If the substitution of - ࣂ for ࣂ in the equation , leaves the
equation unaltered, the curve symmetrical about the initial line.
2. If the power of r are even, the curve is symmetrical
about the pole.
b) Form the table, the value of r, for both positive and negative values of ࣂ
and hence note how r varies with ࣂ . Find in particular the value of ࣂ which gives
r = 0 and r = ∞ .
c) Find tan ߮ . This will indicate the direction of the tangent.
ࣂ that are contained between certain limits.
d) Sometimes by the nature of the equation it is possible to ascertain the value of r and
e) Transform into Cartesian, if necessary and adopt the method given before.
f) Sketch the figure.
PROBLEMS FOR TRACING THE CURVES
1. Astroid : ࢞ ൌ ࢇ ࢉ࢕࢙૜ ࢚ , ࢟ ൌ ࢇ ࢙࢏࢔૜ ࢚ ሺ࢕࢘ ࢞૜ ൅ ࢟૜ ൌ ࢇ૜ )
૛ ૛ ૛
It is symmetrical about the x-axis
Limits |‫ |ݔ‬൑ ܽ
and |‫ |ݕ‬൑ ܽ
The curve lies entirely within the square bounded by the lines ‫ ݔ‬ൌ േܽ , ‫ ݕ‬ൌ േ ܽ

8. Points: we have
when t = 0 or
, when t ൌ
As t increases x Y Portion traced
+ve and +ve and
From 0 to
decreases from increases from From 0 to ∞ A to B
a to 0 0 to a
-ve
ve and
+ ve and
From to π increases
decreases From ∞to 0 B to C
numerically
from a to 0
from 0 to -a
As t increases from π to 2π, we get the reflection of the curve ABC in the x - axis.
π,
,
The values of t > 2π give no new points.
Hence the shape of the curve is as shown in the fig.
Y
-X O X
-Y
Here ox = oy = a

9. It is symmetrical about the y axis. As such we may consider the curve only for
positive values of x or .
Limits: The greatest value of y is 2a and the least value is zero. Therefore the curve
lies entirely between the lines y = 0 and y = 2a.
Points: We have
As increases x Y Portion traced
From 0 to π increases from increases from
From ∞ to 0 0 to A
0 to aπ 0 to 2π
From π to 2π increases from decreases
From 0 to ∞ A to B
a π to 2aπ from 2a to 0
As decreases from 0 to - 2π, we get the reflection of the arch OAB in the y
, y-
axis. Hence the shape of the curve is as in the fig.
Y
X
2. Cardioid: r
Fig.,
Initial Line
A cardioids is symmetrical about the initial line and lies entirely within the circle r
= 2a. Its name has been derived from the Latin word ‘ Kardia’- meaning heart.
Kardia’
Because it is a heart shaped curve.
*********************************************************************
***

10. APPLICATIONS OF CURVE TRACING
I) Length
II) Area
III) Volume
IV) Surface area
Table to find the values: Area, Length ,VolumeThe surface area
Quantity Area (A) Length (S) By revolving about the axis of
rotation to form solid
Coordinate Volume (V) Surface area (SA)
௕ ௕ ௕ ௕
݀‫ݏ‬
system
න ݂ሺ‫ݔ‬ሻ݀‫ݔ‬ න ට1 ൅ ‫ݕ‬ଵ ݀‫ݔ‬
ଶ
න ߨ‫ ݕ‬ଶ ݀‫ݔ‬ න 2ߨ‫ݕ‬ ݀‫ݔ‬
݀‫ݔ‬
Cartesian
௔ ௔ ௔ ௔
form y = f (x)
ൌ ඥ1 ൅ ‫ݕ‬ଵ
ௗ௦
Where
ଶ
ௗ௫
௕
݀‫ݔ‬ ௕ ௕
݀‫ݔ‬ ௕
݀‫ݏ‬
න ݂ሺ‫ݔ‬ሻ ݀‫ ݐ‬න ඥሺ‫ ݔ‬ଵ ሻଶ ൅ ሺ‫ ݕ‬ଵ ሻଶ ݀‫ݐ‬ න ߨ‫ ݕ‬ଶ ݀‫ݐ‬ න 2ߨ‫ݕ‬ ݀‫ݐ‬
݀‫ݐ‬ ݀‫ݐ‬ ݀‫ݐ‬
Parametric
௔ ௔ ௔ ௔
form x= x(t)
y= y(t) Where
ൌ ඥሺ‫ ݔ‬ଵ ሻଶ ൅ ሺ‫ ݕ‬ଵ ሻଶ
ௗ௦
ௗ௧
1 ௕ ௕ ௕
݀‫ݔ‬ ௕
݀‫ݏ‬
න ‫ ݎ‬ଶ ݀ߠ න ට‫ ݎ‬ଶ ൅ ‫ݎ‬ଵଶ ݀ߠ න ߨ‫ݕ‬ ݀ߠଶ
න 2ߨ‫ݕ‬ ݀ߠ
௔ 2 ݀ߠ ݀ߠ
Polar form
௔ ௔ ௔
r = f (ߠ)
Where ൌ
ௗ௦
ௗఏ
ඥ‫ ݎ‬ଶ ൅ ‫ݎ‬ଵଶ
3. Find the entire length of the cardioid ‫ ݎ‬ൌ ܽ ሺ1 ൅ ܿ‫ߠ ݏ݋‬ሻ, Also show that the upper
half is bisected by ߠ ൌ The cardioid is symmetrical about the initial line and for its
గ
ଷ
upper half, increases from 0 to π
Also ‫,݋ݏ݈ܣ‬ ൌ -a sin θ.
ௗ௥
ௗఏ
ௗ௥ ଶ
‫ ׵‬Length of the curve ൌ 2 ‫׬‬଴ ට൤‫ ݎ‬ଶ ൅ ቂ ቃ ൨ dߠ
గ
ௗఏ
2 ‫׬‬଴ ඥሾܽ ሺ1 ൅ ܿ‫ߠ ݏ݋‬ሻଶ ൅ ሺെܽ ‫ߠ݊݅ݏ‬ሻଶ ሿ݀ߠ
గ
2 ‫ ׬‬ඥሺ2ሺ1 ൅ ܿ‫ߠݏ݋‬ሻ݀ߠ
=
గ
଴
=

11. = 4a ‫ ׬‬ඥܿ‫ߠ݀2/ߠݏ݋‬
గ
଴
ୱ୧୬஘/ଶ ஠
= 4a ቚ ቚ
ଵ/ଶ ଴
= 8a (sin π/2 - sin 0) = 8a
‫ ׵‬Length of upper half of the curve of the cruve is 4a. Also length of the arch AP
from 0 to π/3
ഏ ഏ
= 2 ‫ ׬‬య ඥሺ2ሺ1 ൅ ܿ‫ߠݏ݋‬ሻ݀ߠ = 2 ‫ ׬‬య cos . ݀ߠ
ఏ
଴ ଴ ଶ
= 4ܽ|sin ߠ/2|଴
గ/ଷ
= 4ܽ ቂ െ 0ቃ
ଵ
ଶ
= 2a ( half the length of upper half of the cardioids )
Fig.
࢞૜ ൅ ࢟૜ =ࢇ૜
૛ ૛ ૛
1. Find the entire length of the curve
Solution:
The equation to the curve is ‫ ݔ‬య ൅ ‫ ݕ‬య =ܽ య ), the curve is symmetrical about the axis and
మ మ మ
it meets the ‘x’ – axis at x = a
Fig.
If S1 = the length of the curve AB
Then required length is 4S1
ଶ
S = 4S1 = ‫ ׬‬ට1 ൅ ቀ ቁ ݀‫ݔ‬
௔ ௗ௬
଴ ௗ௫
Now,
௬ ଵ/ଷ
‫ ݔ‬య ൅ ‫ ݕ‬య =ܽ య ‫׵‬ ൌ െቀ ቁ
మ మ మ ௗ௬
ௗ௫ ௫
S = 4 ‫ ׬‬ට1 ൅ ቀ ቁ
ଶ/ଷ
݀‫ݔ‬
௔ ௬
଴ ௫
4 ‫׬‬଴ ට ݀‫ݔ‬
௔ ௫ మ/య ା ௬ మ/య
௫ మ/య
=

12. 4 ‫׬‬଴ ට ݀‫ݔ‬
௔ ௔మ/య
௫ మ/య
=
௔
4 ‫׬‬଴ ܽଵ/ଷ ‫ି ݔ‬ଵ/ଷ ݀‫ ݔ‬ൌ 4 ܽଵ/ଷ ቔ ቕ
௔ ௫ మ/య
ଶ/ଷ ଴
=
s = 6a units
Fig.
2. Find the perimeter of cardioid r = a (1+cosθ).
Solution:
The equation to the curve is symmetrical about the initial line.
Fig.
‫ ׵‬The required length of the curve is twice the length of the curve OPA
‫׵‬ ൌ െܽ‫ߠ݊݅ݏ‬
At O, θ = π and at A θ = 0
ௗ௥
ௗఏ
Now, r = a(1+cosθ)
ଶ
૛ ‫ ׬‬ට‫ ݎ‬ଶ ൅ ቀ ቁ ݀ߠ
గ ௗ௥
଴ ௗఏ
s=
૛ ‫ ׬‬ඥaଶ ሺ1 ൅ cosθሻଶ ൅ ܽ ଶ ‫݊݅ݏ‬ଶ ߠ ݀ߠ
గ
଴
s=
2 ‫2 ׬‬acos dθ
஠ ஘
଴ ଶ
s=
ࣂ ࣊
૝ࢇ ቞ ቟
࢙࢏࢔
૛
s= ૚ = 8a units
૛ ૙
Find the area of the ‫ ݔ‬య ൅ ‫ ݕ‬య =ܽ య
మ మ మ
3.
Solution: The parametric equation to the curve ‫ ݔ‬య ൅ ‫ ݕ‬య =ܽ య is given by :
మ మ మ
xൌ ܽ ܿ‫ ݏ݋‬ଷ ߠ , ‫ ݕ‬ൌ ܽ ‫݊݅ݏ‬ଷ ߠ
Area = 4 ‫ݔ݀ ݕ ׬‬ xൌ ܽ ܿ‫ ݏ݋‬ଷ ߠ , ‫ ݕ‬ൌ ܽ ‫݊݅ݏ‬ଷ ߠ
௔
଴
Put
‫ ׵‬dx = -3a cos2θ sinθ dθ when x = 0, θ = π/2 ; when x = a, θ = 0
4‫׬‬ ܽ‫݊݅ݏ‬ଷ ߠሺ-3a cos2θ sinθ dθ
గ/ଶ
଴
A =
12ܽ ଶ ‫׬‬ ‫݊݅ݏ‬ସ ߠ cos2θ dθ
గ/ଶ
଴
=
12ܽ ଶ ቀ . ቁቀ ቁ
ଷ ଵ ଵ గ
ସାଶ ଶାଶ ଶ ଶ
=
Sq. units
ଷగ௔మ
଼
=

13. 4. Find the area of the cardioid r = a (1+cosθ).
‫ ׵‬Total area = 2 ൈ area above the line θ = 0
Solution: The curve is symmetrical about the initial line.
= 2 ‫׬‬଴ ‫ ߠ݀ ݎ‬is the formula
గଵ ଶ
ଶ
A for area in polar curves
= 2 ‫׬‬଴ a ሺ1 ൅ cosθሻଶ ݀ߠ
గଵ ଶ
ଶ
A
ఏ ଶ
= ‫׬‬଴ ܽ ቀ2ܿ‫ ݏ݋‬ଶ ቁ ݀ߠ
గ ଶ
ଶ
A
= 4ܽ ଶ ‫׬‬଴ ܿ‫ ݏ݋‬ସ ݀ߠ
గ ఏ
ଶ
Put θ/2 = t ‫ ߠ݀ ׵‬ൌ 2݀‫ݐ‬
A
4ܽ ଶ ‫׬‬ ܿ‫ ݏ݋‬ସ ‫ݐ݀2 ݐ‬
గ/ଶ
଴
=
ૡࢇ૛ ൈ ൈ
૜ ૚ ࣊
૝ ૛ ૛
=
Sq. units
ଷగ௔మ
ଶ
=
xൌ ܽ ሺ‫ ݐ‬െ ‫ݐ݊݅ݏ‬ሻ, ‫ ݕ‬ൌ ܽ ሺ1 െ ܿ‫ ݐݏ݋‬ሻ, 0 ൑ ‫ ݐ‬൑ 2ߨ and its base.
5. Find the area bounded by an arch of the cycloid
Solution: xൌ ܽ ሺ‫ ݐ‬െ ‫ݐ݊݅ݏ‬ሻ, ‫ ݕ‬ൌ ܽ ሺ1 െ ܿ‫ ݐݏ݋‬ሻ for this arch ‘t’ varies from 0 to 2π.
‫ܽ݁ݎܣ ׵‬ ൌ ‫׬‬௧ୀ଴ ‫ݔ݀ ݕ‬
ଶగ
= ‫ ܽ ׬‬ሺ1 െ ܿ‫ ݐݏ݋‬ሻ ܽ ሺ1 െ ܿ‫ ݐݏ݋‬ሻ ݀‫ݐ‬ since dx = ܽ ሺ1 െ ܿ‫ ݐݏ݋‬ሻ݀‫ݐ‬
ଶగ
௧ୀ଴
= ܽ ૛ ‫ ׬‬ሺ1 െ ܿ‫ ݐݏ݋‬ሻଶ ݀‫ݐ‬
ଶగ
௧ୀ଴
=4ܽ ଶ ‫ ݊݅ݏ ׬‬ସ ݀‫ݐ‬
ଶగ ௧
௧ୀ଴ ଶ
= 8ܽ ଶ ‫݊݅ݏ ׬‬ସ ‫ݐ݀ ݐ‬
గ
௧ୀ଴
= 16ܽ ଶ ‫݊݅ݏ ׬‬ସ ‫ݐ݀ ݐ‬
గ/ଶ
௧ୀ଴
= ૚૟ࢇ૛ . . .
૜ ૚ ࣊
૝ ૛ ૛
= 3ߨܽ ଶ ܵ‫ݏݐ݅݊ݑ .ݍ‬
6. Find the volume generated by revolving the cardiod r = a (1+cosθ) about the
initial line.
Solution: For the curve, varies from 0 to
____________________________________________________________________
Find the volume of the solid obtained by revolving the Astroid x2/3 + y2/3 = a2/3

14. Solution: the equation of the asteroid is x2/3 + y2/3 = a2/3
Volume is obtained by revolving the curve from x = 0 to x = a about x-axis and taking
two times the result.
ܸ ൌ 2 ‫׬‬଴ ߨ‫ ݕ‬ଶ ݀‫ݔ‬
௔
଴
ൌ 2 න ߨሺܽ‫ ݊݅ݏ‬ଷ ߠሻଶ ሺെ3ܽ ܿ‫ߠ݊݅ݏߠ2ݏ݋‬ሻ ݀ߠ
గ
ଶ
ൌ ࢉ࢛࢈࢏ࢉ ࢛࢔࢏࢚࢙
૜૛࣊ࢇ૜
૚૙૞
_____________________________________________________________________
___
Problems for practice:
2. Find the volume of the solid obtained by revolving the cissoid ‫ ݕ‬ଶ ሺ2ܽ െ ‫ ݔ‬ሻ ൌ
1. Find the surface area of r = a (1 - cosθ)
‫ ݔ‬ଷ about its asymptote.
3. Find the length between [0, 2ߨ ] of the curve ‫ ݔ‬ൌ ܽሺߠ ൅ sin ߠሻ, ‫ ݕ‬ൌ ܽሺ1 െ
cos ߠሻ.
____________________________________________________________________
Find the surface area of solid generated by revolving the astroid ‫ ݔ‬య ൅ ‫ ݕ‬య =ܽ య
మ మ మ
about the axis.
Solution: The required surface area is equal to twice the surface area generated by
revolving the part of the astroid in the first quadrant about the axis.
Taking xൌ ܽ ܿ‫ ݏ݋‬ଷ ‫ ݕ , ݐ‬ൌ ܽ ‫݊݅ݏ‬ଷ ‫ ݐ‬we have,
Surface area = 2 ‫׬‬଴ 2ߨ‫ ݔ݀ݕ‬ൌ 4 ‫׬‬଴ ‫ݕ‬ ݀‫ݐ‬
గ/ଶ గ/ଶ ௗ௦
ௗ௧
‫ ݕ‬ටቀ ቁ ൅ ቀ ቁ ݀‫ݐ‬
ௗ௬ ଶ
=4 ‫׬‬଴
గ/ଶ ௗ௫
ௗ௧ ௗ௧
4π ‫׬‬଴ ሺasinଷ tሻ ሼ ሺെ3acos ଶ t sintሻଶ ൅ ሺ3acos ଶ t sintሻଶ ሽ1/2dt
஠/ଶ
=
12a ଶ ‫׬‬଴ sinସ t cos t dt
஠/ଶ
= Put z = sint
ߨܽ ଶ ܵ‫ݏݐ݅݊ݑ .ݍ‬
ଵଶ
ହ
=
7. Find the surface area of the solid generated when the cardioid r = a (1+cosθ)
revolves about the initial line.

15. Solution: The equation to the curve is r = a (1+cosθ). For the upper part of the
curve, θ varies from 0 to π
Put x = r cosθ, y = rsinθ
‫ ׵‬Surface area = ‫ݏ݀ݕ ߨ2 ׬‬
గ
଴
= 2ߨ ‫׬‬଴ ሺ‫ ߠ݊݅ݏ ݎ‬ሻ ݀ߠ
గ ௗ௦
ௗఏ
ଶ
= 2ߨ ‫׬‬଴ ሺ‫ߠ݊݅ݏ ݎ‬ሻට‫ ݎ‬ଶ ൅ ቀ ቁ ݀ߠ
గ ௗ௥
ௗఏ
2ߨ ‫׬‬଴ ܽሺ1 ൅ ܿ‫ߠݏ݋‬ሻ. ‫ ߠ݊݅ݏ‬ඥܽ ଶ ሺ1 ൅ cos ߠሻଶ ൅ ܽ ଶ ‫݊݅ݏ‬ଶ ߠ ݀ߠ
గ
ૈ
=
= 16ૈa2
UNIT IV: VECTOR CALCULUS
Scalar and Vector point functions:
(I) If to each point p(R) of a region E in space there corresponds a definite
scalar denoted by f(R), then f (R) is called a ‘scalar point function’ in E. The
region E so defined is called a scalar field.
Ex: a) The temperature at any instant
b)The density of a body and potential due to gravitational matter.
(II) If to each point p(R) of a region E in space there corresponds a definite
vector denoted by F(R), then it is called the vector point function in E. the
region E so defined is called a vector field.
Ex: a) The velocity of a moving fluid at any instant
b) The gravitational intensity of force.
Note: Differentiation of vector point functions follows the same rules as those
of ordinary calculus.
If F (x,y,z) be a vector point function then

16. ݀‫ݖ߲ ܨ߲ ݕ߲ ܨ߲ ݔ߲ ܨ߲ ܨ‬
ൌ ൅ ൅
݀‫ݐ߲ ݖ߲ ݐ߲ ݕ߲ ݐ߲ ݔ߲ ݐ‬
݀‫ݖ݀ ܨ߲ ݕ݀ ܨ߲ ݔ݀ ܨ߲ ܨ‬
ൌ ൅ ൅
݀‫ݐ݀ ݖ߲ ݐ݀ ݕ߲ ݐ݀ ݔ߲ ݐ‬
߲‫ܨ‬ ߲‫ܨ‬ ߲‫ܨ‬
݀‫ ܨ‬ൌ ݀‫ ݔ‬൅ ݀‫ ݕ‬൅ ݀‫ݖ‬
߲‫ݔ‬ ߲‫ݕ‬ ߲‫ݖ‬
߲ ߲ ߲
݀‫ ܨ‬ൌ ൬ ݀‫ ݔ‬൅ ݀‫ ݕ‬൅ ݀‫ݖ‬൰ ‫ܨ‬
߲‫ݔ‬ ߲‫ݕ‬ ߲‫ݖ‬
( 1)
Vector operator del ( ∇ )
The operator is of the form
∇ ൌ ݅ ൅݆ ൅݇
డ డ డ
డ௫ డ௬ డ௭
GRADIENT, DIVERGENCE, CURL (G D C)
Gradient of the scalar point function:
It is the vector point function f defined as the gradient of the scalar point
function f and is written as grad f, then
grad f = ∇f = ቀ݅ ൅݆ ൅݇ ቁ݂
డ డ డ
డ௫ డ௬ డ௭
߲݂ ߲݂ ߲݂
ൌ݅ ൅݆ ൅݇
߲‫ݔ‬ ߲‫ݕ‬ ߲‫ݖ‬
DIVERGENCE OF A VECTOR POINT FUNCTION
The divergence of a continuously differentiable vector point function F(div F)is
defined by the equation
߲‫ܨ‬ ߲‫ܨ‬ ߲‫ܨ‬
݀݅‫ ܨ ݒ‬ൌ ∇. ‫ ܨ‬ൌ ݅ ൅݆ ൅݇
߲‫ݔ‬ ߲‫ݕ‬ ߲‫ݖ‬
‫ ܨ ݂ܫ‬ൌ ݂ ݅ ൅ ‫ ݆ ׎‬൅ ߮ ݇ ‫݄݊݁ݐ‬

17. ߲ ߲ ߲
݀݅‫ ܨ ݒ‬ൌ ∇. ‫ ܨ‬ൌ ൬݅ ൅݆ ൅ ݇ ൰ · ሺ݂ ݅ ൅ ‫ ݆ ׎‬൅ ߮ ݇ሻ
߲‫ݔ‬ ߲‫ݕ‬ ߲‫ݖ‬
߲݂ ߲‫߲߮ ׎‬
∇·‫ ܨ‬ൌ ൅ ൅
߲‫ݖ߲ ݕ߲ ݔ‬
CURL OF A VECTOR POINT FUNCTION
The curl of a continuously differentiable vector point function F is defined by
the equation
߲‫ܨ‬ ߲‫ܨ‬ ߲‫ܨ‬
curl F ൌ ∇ ൈ ‫ ܨ‬ൌ ݅ ൈ ൅݆ൈ ൅݇ൈ
߲‫ݔ‬ ߲‫ݕ‬ ߲‫ݖ‬
߲ ߲ ߲
curl F ൌ ∇ ൈ ‫ ܨ‬ൌ ൬݅ ൅݆ ൅ ݇ ൰ ൈ ሺ݂݅ ൅ ‫ ݆׎‬൅ ߮݇ሻ
߲‫ݔ‬ ߲‫ݕ‬ ߲‫ݖ‬
݅ ݆ ݇
߲ ߲ ߲
curl F ൌ ∇ ൈ ‫ ܨ‬ൌ ተ ተ
߲‫ݔ‬ ߲‫ݕ‬ ߲‫ݖ‬
݂ ‫׎‬ ߮
߲߮ ߲‫׎‬ ߲߮ ߲݂ ߲߮ ߲݂
∇ൈ‫ ܨ‬ൌ ݅൬ െ ൰െ݆൬ െ ൰൅݇൬ െ ൰
߲‫ݖ߲ ݕ‬ ߲‫ݖ߲ ݔ‬ ߲‫ݖ߲ ݔ‬
∇݂ ܽ݊݀ ∇ ൈ ݂
∇·‫ܨ‬
DEL APPLIED TWICE TO POINT FUNCTIONS
Le being vector point functions, we can form their divergence and curl,
whereas being a scalar point function, we can have its gradient only. Then we
have Five formulas:
߲ଶ݂ ߲ଶ݂ ߲ଶ ݂
div grad f ൌ ∇ ݂ ൌ ଶ ൅ ଶ ൅ ଶ
ଶ
߲‫ݔ‬ ߲‫ݕ‬ ߲‫ݖ‬
ܿ‫ ݂ ݀ܽݎ݃ ݈ݎݑ‬ൌ ∇ ൈ ∇݂ ൌ 0
3. ݀݅‫ ܨ ݈ݎݑܿ ݒ‬ൌ ∇ · ∇ ൈ ‫ ܨ‬ൌ 0 ݀݅‫ ܨ ݈ݎݑܿ ݒ‬ൌ ∇ · ∇ ൈ ‫ ܨ‬ൌ 0
ܿ‫ ܨ ݈ݎݑܿ ݈ݎݑ‬ൌ ݃‫ ܨ ݒ݅݀ ݀ܽݎ‬െ ∇ଶ F

18. ∇ ൈ ሺ∇ ൈ ‫ܨ‬ሻ ൌ ∇ሺ∇ · ‫ܨ‬ሻ െ ∇ଶ F
݃‫ ܨ ݒ݅݀ ݀ܽݎ‬ൌ ܿ‫ ܨ ݈ݎݑܿ ݈ݎݑ‬൅ ∇ଶ F
∇ሺ∇ · ‫ܨ‬ሻ ൌ ∇ ൈ ሺ∇ ൈ ‫ܨ‬ሻ ൅ ∇ଶ F
PROOF:
સ ૛ ܎ ൌ સ · સ܎ ൌ સ · ቀ࢏ ൅࢐ ൅ ࢑ ቁ܎
ࣔ ࣔ ࣔ
ࣔ࢞ ࣔ࢟ ࣔࢠ
(I) To prove that
߲݂ ߲݂ ߲݂
ൌ ∇ · ൬݅ ൅݆ ൅݇ ൰
߲‫ݔ‬ ߲‫ݕ‬ ߲‫ݖ‬
߲ ߲ ߲ ߲݂ ߲݂ ߲݂
ൌ curl grad f ൌ ∇ ൈ ∇f ൌ ൬݅ ൅݆ ൅ ݇ ൰ ൈ ൬݅ ൅݆ ൅݇ ൰
߲‫ݔ‬ ߲‫ݕ‬ ߲‫ݖ‬ ߲‫ݔ‬ ߲‫ݕ‬ ߲‫ݖ‬
߲ଶ݂ ߲ଶ݂ ߲ଶ ݂
ൌ ଶ൅ ଶ൅ ଶ
߲‫ݔ‬ ߲‫ݕ‬ ߲‫ݖ‬
߲ଶ ߲ଶ ߲ଶ
ൌ ቆ ଶ ൅ ଶ ൅ ଶቇ ݂
߲‫ݔ‬ ߲‫ݕ‬ ߲‫ݖ‬
ൌ ∇ଶ f
߲ଶ ߲ଶ ߲ଶ
‫∇ ݁ݎ݄݁ݓ‬ଶ ൌ ൅ ଶ ൅ ଶ ݅‫ݎ݋ݐܽݎ݌݋ ݈݊ܽ݅ܿܽ݌ܽܮ ݈݈݀݁ܽܿ ݏ‬
߲‫ ݔ‬ଶ ߲‫ݕ‬ ߲‫ݖ‬
ܽ݊݀ ∇ଶ f ൌ 0 ݅‫݊݋݅ݐܽݑݍ݁ ݏ ′ ݈݁ܿܽ݌ܽܮ ݈݈݀݁ܽܿ ݏ‬
(II) To prove that ࢉ࢛࢘࢒ ࢍ࢘ࢇࢊ ࢌ ൌ ૙
߲ ߲ ߲ ߲݂ ߲݂ ߲݂
curl grad f ൌ ‫ ׏‬ൈ ‫׏‬f ൌ ൬݅ ൅݆ ൅ ݇ ൰ ൈ ൬݅ ൅݆ ൅݇ ൰
߲‫ݔ‬ ߲‫ݕ‬ ߲‫ݖ‬ ߲‫ݔ‬ ߲‫ݕ‬ ߲‫ݖ‬
݅ ݆ ݇
߲ ߲ ߲
ተ ተ
ൌ ߲‫ݔ‬ ߲‫ݕ‬ ߲‫ݖ‬
ተ߲݂ ߲݂ ߲݂ተ
߲‫ݔ‬ ߲‫ݕ‬ ߲‫ݖ‬

19. ߲ଶ݂ ߲ଶ݂ ߲ଶ݂ ߲ଶ݂ ߲ଶ݂ ߲ଶ݂
ൌ iቆ െ ቇ െjቆ െ ቇ ൅ kቆ െ ቇ
߲‫ݖ߲ ݕ߲ ݖ߲ ݕ‬ ߲‫ݖ߲ ݔ߲ ݖ߲ ݔ‬ ߲‫ݕ߲ ݔ߲ ݕ߲ ݔ‬
‫ ׏‬ൈ ‫׏‬f ൌ 0
(III) To prove thatસ · સ ൈ ࡲ ൌ ቀ∑ ࢏ ቁ · ቀ࢏ ൈ ࣔ࢞ ൅ ࢐ ൈ ࣔ࢟ ൅ ࢑ ൈ ࣔࢠ ቁ
ࣔ ࣔࡲ ࣔࡲ ࣔࡲ
ࣔ࢞
ൌ ∑ ݅ · ቀ݅ ൈ ൅݆ൈ ൅݇ൈ ቁ
డమ ி డమ ி డమ ி
డ௫ మ డ௬ డ௫ డ௭ డ௫
ൌ ∑ ቀ݅ ൈ ݅ · ൅݅ൈ݆· ൅݅ൈ݇· ቁ
డమ ி డమ ி డమ ி
డ௫ మ డ௫ డ௬ డ௫ డ௭
ൌ ∑ ቀ݇ · െ݆· ቁ
డమ ி డమ ி
డ௫ డ௬ డ௫ డ௭
ൌ0
(IV) To prove that સ ൈ સ ൈ ࡲ ൌ ቀ∑ ࢏ ቁ ൈ ቀ࢏ ൈ ࣔ࢞ ൅ ࢐ ൈ ࣔ࢟ ൅ ࢑ ൈ ࣔࢠ ቁ
ࣔ ࣔࡲ ࣔࡲ ࣔࡲ
ࣔ࢞
߲ଶ‫ܨ‬ ߲ଶ‫ܨ‬ ߲ଶ‫ܨ‬
ൌ ෍ ݅ ൈ ቆ݅ ൈ ൅݆ൈ ൅݇ൈ ቇ
߲‫ ݔ‬ଶ ߲‫ݕ߲ ݔ‬ ߲‫ݖ߲ ݔ‬
߲ଶ‫ܨ‬ ߲ଶ‫ܨ‬ ߲ଶ ‫ܨ‬ ߲ଶ ‫ܨ‬
ൌ ෍ ቈቆ݅ · ଶ ቇ ݅ െ ሺ݅ · ݅ሻ ଶ ቉ ൅ ቈቆ݅ · ቇ ݆ െ ሺ݅ · ݆ሻ ቉
߲‫ݔ‬ ߲‫ݔ‬ ߲‫ݕ߲ ݔ‬ ߲‫ݕ߲ ݔ‬
߲ଶ‫ܨ‬ ߲ଶ‫ܨ‬
൅ ቈቆ݅ · ቇ ݇ െ ሺ݅ · ݇ሻ ቉
߲‫ݖ߲ ݔ‬ ߲‫ݖ߲ ݔ‬
߲ଶ‫ܨ‬ ߲ଶ‫ܨ‬ ߲ଶ‫ܨ‬ ߲ଶ‫ܨ‬
ൌ ෍݅ · ଶ ݅ ൅ ݅ · ݆൅ ݅· ݇െ෍ ଶ
߲‫ݔ‬ ߲‫ݕ߲ ݔ‬ ߲‫ݖ߲ ݔ‬ ߲‫ݔ‬
߲ ߲‫ܨ‬ ߲‫ܨ‬ ߲‫ܨ‬ ߲ଶ‫ܨ‬
ൌ ෍ ݅ ൬݅ · ൅݆· ൅݇· ൰െ෍ ଶ
߲‫ݔ‬ ߲‫ݔ‬ ߲‫ݕ‬ ߲‫ݖ‬ ߲‫ݔ‬
ൌ ‫׏‬ሺ‫ · ׏‬Fሻ െ ‫׏‬ଶ F
curl curl F ൌ ‫׏‬ሺ‫ · ׏‬Fሻ െ ‫׏‬ଶ F
(V) To prove that
We have by (IV) which implies ‫׏‬ሺ‫ · ׏‬Fሻ ൌ ܿ‫ ܨ ݈ݎݑܿ ݈ݎݑ‬൅ ‫׏‬ଶ F
‫׏‬ሺ‫ · ׏‬Fሻ ൌ ‫ ׏‬ൈ ሺ‫ ׏‬ൈ ‫ܨ‬ሻ ൅ ‫׏‬ଶ F

20. ‫ ݔ‬ൌ ‫ ݐ‬ଷ ൅ 1, ‫ ݕ‬ൌ ‫ ݐ‬ଶ , ‫ ݖ‬ൌ ‫ ݐ‬൅ 5
݅൅3݆൅2݇
.
ࡼ࢘࢕࢈࢒ࢋ࢓: ݄ܵ‫ ݐ݄ܽݐ ݓ݋‬સ ଶ r ୬ ൌ nሺn ൅ 1ሻr ୬ିଶ
‫׏ :ܖܗܑܜܝܔܗ܁‬ଶ ሺr ୬ ሻ ൌ ‫ · ׏‬ሺ‫׏‬r ୬ ሻ
R
ൌ ‫ · ׏‬൬n r ୬ିଵ ൰ ൌ n‫ · ׏‬ሺr ୬ିଶ Rሻ ൌ nሾሺ‫׏‬r ୬ିଶ ሻ. R ൅ r ୬ିଶ ሺ‫ · ׏‬Rሻሿ
r
R
ൌ n ൤ሺn െ 2ሻr ୬ିଷ · R ൅ r ୬ିଶ ሺ3ሻ൨
r
ൌ nሾሺn െ 2ሻr ୬ିସ ሺr ଶ ሻ ൅ 3r ୬ିଶ ሿ
ൌ nሺn ൅ 1ሻr ୬ିଶ
Otherwise: ‫׏‬ଶ ሺr ୬ ሻ ൌ ൅ ൅
பమ ሺ୰౤ ሻ பమ ሺ୰౤ ሻ பమ ሺ୰౤ ሻ
ப୶మ ப୷మ ப୸మ
Now ൌ nr ୬ିଵ ൌ nr ୬ିଵ ൌ nr ୬ିଶ x
பሺ୰౤ ሻ ப୰ ୶
ப୶ ப୶ ୰
ሺ∂^2 ሺr^n ሻሻ/ሺ∂x^2 ሻ ൌ nሾr^ሺn െ 2ሻ ൅ ሺn െ 2ሻ r^ሺn െ 3ሻ ∂r/
∂x xሿ ൌ nሾr^ሺn െ 2ሻ ൅ ሺn െ 2ሻ r^ሺn െ 3ሻ x/r xሿ
ൌ nሾr ୬ିଶ ൅ ሺn െ 2ሻr ୬ିସ x ଶ ሿ … … … … … . ሺ1ሻ
SImilarly, ൌ nሾr ୬ିଶ ൅ ሺn െ 2ሻr ୬ିସ y ଶ ሿ … … … … . . ሺ2ሻ
பమ ሺ୰౤ ሻ
ப୷మ
ൌ nሾr ୬ିଶ ൅ ሺn െ 2ሻr ୬ିସ z ଶ ሿ … … … … … ሺ3ሻ
பమ ሺ୰౤ ሻ
ப୸మ
Adding equations ሺ1ሻ, ሺ2ሻand ሺ3ሻ, gives
‫׏‬ଶ ሺr ୬ ሻ ൌ nሾ3r ୬ିଶ ൅ ሺn െ 2ሻr ୬ିସ ሺx ଶ ൅ y ଶ ൅ z ଶ ሻሿ
ൌ nሾ3r ୬ିଶ ൅ ሺn െ 2ሻr ୬ିସ r ଶ ሿ ൌ nሺn ൅ 1ሻr ୬ିଶ
In particular ‫׏‬ଶ ቀ ቁ ൌ 0
ଵ
୰

21. Ex: A particle moves along the curve ‫ ݔ‬ൌ ‫ ݐ‬ଷ ൅ 1, ‫ ݕ‬ൌ ‫ ݐ‬ଶ , ‫ ݖ‬ൌ ‫ ݐ‬൅ 5 find the
components of velocity and acceleration at t=2 in the direction of ݅ ൅ 3 ݆ ൅ 2 ݇
Solution: ݄ܶ݁ ‫,ݏ݅ ݐ ݁݉݅ݐ ݐܽ ݈݁ܿ݅ݐݎܽ݌ ݄݁ݐ ݂݋ ݎ݋ݐܿ݁ݒ ݊݋݅ݐ݅ݏ݋݌‬

22. ܴ ൌ‫݅ݔ‬൅‫݆ݕ‬൅‫݇ݖ‬
ሬԦ
ܴ ൌ ሺ‫ ݐ‬ଷ ൅ 1ሻ݅ ൅ ‫ ݐ‬ଶ ݆ ൅ ሺ‫ ݐ‬൅ 5ሻ ݇
ሬԦ
݄ܶ݁ ‫ ܸ ݕݐ݅ܿ݋݈݁ݒ‬ൌ ݅‫ݕܾ ݊݁ݒ݅݃ ݏ‬
ௗோ
ሬԦ
ௗ௧
ሬԦ
ܸൌ
ሬԦ ௗோ ൌ 3‫ ݐ‬ଶ ݅ ൅ 2 ‫ ݆ ݐ‬൅ ݇
ሬԦ
ௗ௧
݄ܶ݁ ݈ܽܿܿ݁݁‫ ܽ ݊݋݅ݐܽݎ‬ൌ ൌ ݅‫ݕܾ ݊݁ݒ݅݃ ݏ‬
ௗ௏
ሬԦ ௗమோ
ሬԦ
ௗ௧ ௗ௧ మ
Ԧ
ܽൌ ൌ6‫݅ݐ‬൅2݆
ௗమோ
ሬԦ
ௗ௧ మ
Ԧ
݄ܶ݁݊ ܽ‫ ݐ ݁݉݅ݐ ݐ‬ൌ 2, ܸ ܽ݊݀ ܽ ܽ‫ݕܾ ݊݁ݒ݅݃ ݁ݎ‬
ሬԦ Ԧ
ܸൌ
ሬԦ ௗோ ൌ 12 ݅ ൅ 4 ݆ ൅ ݇
ሬԦ
ௗ௧
ܽൌ ൌ 12 ݅ ൅ 2 ݆
ௗమ ோ
ሬԦ
ௗ௧ మ
Ԧ
ܸ݈݁‫ ݅ ݂݋ ݊݋݅ݐܿ݁ݎ݅݀ ݄݁ݐ ݊݅ ݎ݋ݐܿ݁ݒ ݕݐ݅ܿ݋‬൅ 3݆ ൅ 2݇ ݅‫ݏ‬
ሬԦ ௜ାଷ ௝ାଶ ௞ቁ ൌ ሺ12 ݅ ൅ 4 ݆ ൅ ݇ሻ · ቀ௜ାଷ ௝ାଶ ௞ቁ ൌ
ൌ ܸ·ቀ
ଶ଺
√ଵସ √ଵସ √ଵସ
‫ ݅ ݂݋ ݊݋݅ݐܿ݁ݎ݅݀ ݄݁ݐ ݊݅ ݊݋݅ݐܽݎ݈݁݁ܿܿܣ‬൅ 3݆ ൅ 2݇ ݅‫ݏ‬
ൌܽ·ቀ ቁ ൌ ሺ12 ݅ ൅ 2 ݆ሻ · ቀ ቁൌ
௜ାଷ ௝ାଶ ௞ ௜ାଷ ௝ାଶ ௞ ଵ଼
√ଵସ √ଵସ √ଵସ
Ԧ
ࡼ࢘࢕࢈࢒ࢋ࢓: ‫ ݕݔ ݂݁ܿܽݎݑݏ ݄݁ݐ ݋ݐ ݈ܽ݉ݎ݋݊ ݎ݋ݐܿ݁ݒ ݐ݅݊ݑ ݄݁ݐ ݀݊݅ܨ‬ଷ ‫ ݖ‬ଶ ൌ
4 ܽ‫ ݐ݊݅݋݌ ݄݁ݐ ݐ‬ሺെ1, െ1,2ሻ
ࡿ࢕࢒࢛࢚࢏࢕࢔: ܸ݁ܿ‫׏ ݏ݅ ݂݁ܿܽݎݑݏ ݊݁ݒ݅݃ ݄݁ݐ ݋ݐ ݈ܽ݉ݎ݋݊ ݎ݋ݐ‬ሺ‫ ݕݔ‬ଷ ‫ ݖ‬ଶ ሻ
‫׏‬ሺ‫ ݕݔ‬ଷ ‫ ݖ‬ଶ ሻ ൌ ቀ݅ ൅݆ ൅݇ ቁ ሺ‫ ݕݔ‬ଷ ‫ ݖ‬ଶ ሻ
డ డ డ
డ௫ డ௬ డ௭
ൌ ‫ ݕ‬ଷ ‫ ݖ‬ଶ ݅ ൅ 3‫ ݕݔ‬ଶ ‫ ݖ‬ଶ ݆ ൅ 2‫ ݕݔ‬ଷ ‫݇ ݖ‬
‫ ݐܣ‬ሺെ1, െ1,2ሻ
ൌ 4 ݅ െ 12 ݆ ൅ 4 ݇
‫ݏ݅ ݂݁ܿܽݎݑݏ ݄݁ݐ ݋ݐ ݈ܽ݉ݎ݋݊ ݐ݅݊ݑ ݀݁ݎ݅ݏ݁݀ ݄݁ݐ ݁ܿ݊݁ܪ‬
ൌ ൌ ሺ4݅ െ 12݆ ൅ 4݇ሻ
ସ ௜ିଵଶ௝ାସ௞ ଵ
√ଵ଺ାଵସସାଵ଺ √ଵ଻଺

23. ࡼ࢘࢕࢈࢒ࢋ࢓: ‫݂ ݂݋ ݁ݒ݅ݐܽݒ݅ݎ݁݀ ݈ܽ݊݋݅ݐܿ݁ݎ݅݀ ݄݁ݐ ݀݊݅ܨ‬ሺ‫ݖ ,ݕ ,ݔ‬ሻ ൌ ‫ ݕݔ‬ଶ ൅
‫ ݖݕ‬ଷ ܽ‫ ݐ݊݅݋݌ ݄݁ݐ ݐ‬ሺ2, െ1,1ሻ ݅݊ ‫ ݅ ݂݋ ݊݋݅ݐܿ݁ݎ݅݀ ݄݁ݐ‬൅ 2݆ ൅ 2݇
ࡿ࢕࢒࢛࢚࢏࢕࢔: ܸ݁ܿ‫݂ ݂݁ܿܽݎݑݏ ݄݁ݐ ݋ݐ ݈ܽ݉ݎ݋݊ ݎ݋ݐ‬ሺ‫ݖ ,ݕ ,ݔ‬ሻ ݅‫݂׏ ݏ‬
i. e. ‫ ݂׏‬ൌ ቀ݅ ൅݆ ൅݇ ቁ ሺ‫ ݕݔ‬ଶ ൅ ‫ ݖݕ‬ଷ ሻ
డ డ డ
డ௫ డ௬ డ௭
‫ ݂׏‬ൌ ‫ ݕ‬ଶ ݅ ൅ ሺ2‫ ݕݔ‬൅ ‫ ݖ‬ଷ ሻ ݆ ൅ ሺ3‫ ݖݕ‬ଶ ሻ ݇
‫ ݐܣ‬ሺ2, െ1,1ሻ
‫ ݂׏‬ൌ ݅ െ 3 ݆ െ 3 ݇
‫ ݅ ݂݋ ݊݋݅ݐܿ݁ݎ݅݀ ݄݁ݐ ݊݅ ݁ݒ݅ݐܽݒ݅ݎ݁݀ ݈ܽ݊݋݅ݐܿ݁ݎ݅ܦ‬൅ 2݆ ൅ 2݇ ݅‫ݏ‬
ൌ ሺ‫݂׏‬ሻሺଶ,ିଵ,ଵሻ · ൌ ሺ݅ െ 3݆ െ 3݇ሻ ·
௜ାଶ௝ାଶ௞ ௜ାଶ௝ାଶ௞
√ଵାସାସ √ଽ
ൌ ൌ
ଵି଺ି଺ ିଵଵ
ଷ ଷ
‫׏‬ଶ ሺr ୬ ሻ ൌ nሺn ൅ 1ሻr ୬ିଵ
‫׏‬ଶ ሺr ୬ ሻ ൌ ‫׏‬ሺ‫׏‬r ୬ ሻ
ൌ
ࡼ࢘࢕࢈࢒ࢋ࢓: ‫ݎ݋ݐܿ݁ݒ ݄݁ݐ ݐ݄ܽݐ ݄ܿݑݏ ܿ ,ܾ ,ܽ ݏݐ݊ܽݐݏ݊݋ܿ ݄݁ݐ ݀݊݅ܨ‬
‫ ܨ‬ൌ ሺ‫ ݔ‬൅ ‫ ݕ‬൅ ܽ‫ݖ‬ሻଓ̂ ൅ ሺ‫ ݔ‬൅ ܿ ‫ ݕ‬൅ 2‫ݖ‬ሻ݇ ൅ ሺܾ‫ ݔ‬൅ 2‫ ݕ‬െ ‫ݖ‬ሻଔ̂ ݅‫݈ܽ݊݋݅ݐܽݐ݋ݎݎ݅ ݏ‬
Ԧ ෠
ࡿ࢕࢒࢛࢚࢏࢕࢔:
‫ ܨ ݊݁ݒ݅ܩ‬ൌ ሺ‫ ݔ‬൅ ‫ ݕ‬൅ ܽ‫ݖ‬ሻଓ̂ ൅ ሺ‫ ݔ‬൅ ܿ ‫ ݕ‬൅ 2‫ݖ‬ሻ݇ ൅ ሺܾ‫ ݔ‬൅ 2‫ ݕ‬െ ‫ݖ‬ሻଔ̂
Ԧ ෠
݅‫݈ܽ݊݋݅ݐܽݐ݋ݎݎ݅ ݏ‬
ܵ݅݊ܿ݁ ‫݈ܽ݊݋݅ݐܽݐ݋ݎݎ݅ ݏ݅ ݈݂݀݁݅ ݎ݋ݐܿ݁ݒ ݄݁ݐ‬
݄ܶ݁‫ ׏ ݁ݎ݋݂݁ݎ‬ൈ ‫ ܨ‬ൌ 0
Ԧ
ଓ̂ ଔ̂ ݇
෠
‫ ܨ ݈ݎݑܥ‬ൌ ተ ተ
డ డ డ
డ௫ డ௬ డ௭
Ԧ
‫ ݔ‬൅ ‫ ݕ‬൅ ܽ‫ݖ‬ ܾ‫ ݔ‬൅ 2‫ ݕ‬െ ‫ݖ‬ ‫ ݔ‬൅ ܿ‫ ݕ‬൅ 2‫ݖ‬
ൌ ሺܿ ൅ 1ሻଓ̂ െ ሺ1 െ ܽሻଔ̂ ൅ ሺܾ െ 1ሻ݇
෠
݅. ݁. , ሺܿ ൅ 1ሻଓ̂ െ ሺ1 െ ܽሻଔ̂ ൅ ሺܾ െ 1ሻ݇ ൌ 0
෣
݄ܶ݅‫ ܿ ,݄݊݁ݓ ݕ݈݊݋ ݈ܾ݁݅ݏݏ݋݌ ݏ݅ ݏ‬െ 1 ൌ 0, 1 െ ܽ ൌ 0, ܾ െ 1 ൌ 0
‫ ܽ ݏ݈݁݅݌݉݅ ݄݄ܿ݅ݓ‬ൌ 1, ܾ ൌ 1, ܿ ൌ 1
Orthogonal curvilinear co-ordinates

24. Let the rectangular co-ordinates (x,y,z) of Any point be expressed as
co ordinates
function of (u,v,w),
So that x = x(u,v,w),y = y(u,v,w),z = z(u,v,w) …..(1)
Suppose that (1) can be solved for u,v,w in terms of x,y,z
i,e u = (x,y,z), v = v(x,y,z),w = w(x,y,z) ….(2)
We assume that the functions in (1) and (2) are single valued functions
and have continuous partial derivatives so that the correspondence between
(x,y,z) and (u,v,w) is unique. Then (u,v,w) are called curvilinear co-ordinates
curvilinear co
of (x,y,z). Each of u,v,w has a level of surface through an arbitrary point .
t bitrary
The surface are called co
co-ordinate
surface through
Each pair of these co-ordinate surface intersects In curves called the co
co ordinate co-
ordinate curves. The curve of intersection of will be called the w
w-curve, for
only w changes along this curve. Similarly we define u and v-curves.
v curves.
In vector notation, (1) can be written as
R = x(u,v,w)I + y(u,v,w)J + z(u,v,w)K

28. The co-ordinate curves for ρ are rays perpendicular to the Z-axis; for ф
axis;
horizontal circles with centers on the Z-axis; for z lines parallels to the Z-axis.
Z axis; Z
x = ρ cos ф,

29. y = ρ sin ф, z=z
So that scale factors are h1=1, h2 = ρ, h3= 1. Also the volume element
dV=ρ dρ dф dz.
2) Spherical polar co-
-ordinates:
Let p(x,y,z) be any point whose projection on the xy-plane is Q(x,y) .
plane
Then the Spherical polar co-
-ordinates of p are such that r = op,
.
The level surfaces are respectively spheres
about O, cones about the Z-axis with vertex at O and planes through the Z-axis.
axis Z
The co-ordinate curves for r are rays from the origin; for θ, vertical circles with
ordinate ,
centre at O (called meridians); for ф, horizontal circles with centres on the Z
, Z-
axis.
x = OQ cosф
= OP cos(90-θ)cos
)cosф
= r sinθ cosф,
y = OQ sinф = r sinθ sinф
z = r cosθ
So that the scale factors are
Also the volume element