1) The document discusses formulas for calculating area, volume, fluid pressure, and work based on the cross-sectional lengths and areas of regions and solids. 2) It provides examples of calculating the area of regions bounded by functions, the volume of solids of revolution, fluid pressure on a plate, and work needed to pump water. 3) The key concepts are using integrals to calculate quantities by summing cross-sectional lengths or areas and defining these lengths and areas based on the geometry of regions and solids.

1. Formulas Based on Cross Sections

2. Formulas Based on Cross Sections
The cross–sectional lengths L(x) of a region and
the cross–sectional areas A(x) of a solid are the bases for
many of the formulas for area, volume, work/H2O pressure.
Here is summary of the methods. Here are the basic algebra
and geometry:
R
L
Horizontal distance/length
h = Right – Left
Distance h = R – L
x
y
H
L
Distance
v = H – L
Vertical distance/length
h = High – Low

3. Formulas Based on Cross Sections
The cross–sectional lengths L(x) of a region and
the cross–sectional areas A(x) of a solid are the bases for
many of the formulas for area, volume, work/H2O pressure.
Here is summary of the methods. Here are the basic algebra
and geometry:
Vertical vs horizontal distance/length
to the axes from
a point (x, y) on
the graph F(x, y) = 0
F(x, y) = 0
R
L
Horizontal distance/length
h = Right – Left
Distance h = R – L
x
y
H
L
Distance
v = H – L
Vertical distance/length
h = High – Low

4. Area = integral of length L
dx
L(x)
dA = L(x) dx
Σ dA → ∫
x=a
b
L(x) dx
= Area of A
A FTC
Type I Region Area Formula: L(x) = Vertical distances
∫
b
L(x) dx
a
=
∫
b
f(x) – g(x) dx
a
Area

5. Type Il Region Area Formula: L(y) = Horizontal Distances
∫
d
L(y) dy
c
=
∫
d
f(y) – g(y) dy
c
L(y) dA = L(y) dy
Σ dA → ∫
x=a
b
L(y) dy
= Area of A
FTC
A
Area = integral of length L
Area

6. Example A. (Type I region: High – Low)
Find the area bounded by y = –x2 + 2x and y = x2
Area
Set the equations equal to
find the intersection points.
x2 = –x2 + 2x so x = 0, 1.
A = ∫
x=0
1
(–x2 + 2x) – x2 dx = 1
3
y = –x2 + 2x
y = x2
1
y = x – 2 so x = y + 2
y = √x
so x = y2
L(y) = y + 2 – y2
y = 2
Solve for x for the border functions.
The right border is x = y + 2,
the left border is x = y2 and that
L(y) = (y + 2) – y2 with y = 0 to y = 2.
So area A = y + 2 – y2 dy
∫
y = 0
2
=
10
3
B. (Type II region: Right – Left)
Find the area bounded by y = √x and y = x – 2.

7. dV = A(x) dx
V A(x)
dx
Σ dV → ∫
x=a
b
A(x) dx
= Volume of V
FTC
Volume = integral of area
Volume of revolution–solid
Solids of Revolution
∫
x=a
b
dx
V = π [f (x)]2
Cross–section–area A(x) = π [f (x)]2:

8. dV = A(y) dy
V
A(y)
dy
Σ dV → ∫
x=a
b
A(y) dy
= Volume of V
FTC
∫y=a
dy
π [f (y)]2
V =
Solids of Revolution
Volume = integral of area
Volume of revolution–solid
Cross–section–area A(x) = π [f (y)]2:
b

9. Solids of Revolution
Example C. Find the volume of revolution around the
x axis formed by the curve y = √x from x = 0 to x = 2.
The cross–sections are discs with
radius r = √x so A(x) = π (√x)2.
∫
x=0
2
dx
So the volume V = πx = 2π
∫
y=1
π [(1/2 + y–1/2)2 – 1/2 2] dy
4
So the volume V =
D. Find the volume of revolution around
the axis x = –½ formed by the template as shown.
The inner radius r = ½,
outer radius R = ½ + y –1/2.
2
=19π

10. ∫
b
ρ D(x) L(x)dx
F =
D(x)
L(x)
x=a
Fluid Pressures F
ρ – fluid density
dA

11. ∫
b
ρ D(x) L(x)dx
F =
D(x)
L(x)
x=a
Fluid Pressures F
ρ – fluid density
Work (Pumping H2O)
W = ∫x=a
b
D(x) A(x)dx
ρ
A(x)
D(x)
dA
dV

12. Place the x-axis as shown
so L(x) is 2100 – x2
and the distance D(x) = x.
0
x
10
100 – x2
So the pressure
P = ρ 2x100 – x2 dx = 41,600 lb
∫
x=0
10
Example E. Find the amount of water force on a
semi–disc plate with 10-ft radius as shown.
Fluid Pressures
A(x) = π(3x/4) 2 =9πx2/16 and
distance D(x) = (18 – x).
So the total work W = ∫
x=0
8
9πx2
(18 – x)dx =1152πρ
16
ρ
F. Find the work W needed to pump all the water
over the height as shown.