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# Volumes with Known Cross Sections Worksheet

Volumes with Known Cross Sections Worksheet
• Page 1
1.
What is the volume of a body bounded by the planes $x$ = 2, $x$ = 7 whose area of cross section perpendicular to $x$-axis is inversely proportional to the square of the distance of the section from the origin and the area of the cross section at $x$ = 4 is 13 square units?
 a. 41.6 b. 77.29 c. 74.29 d. 38.6

#### Solution:

The area of cross section perpendicular to x-axis of the solid = A(x) 1x2 where x is the distance of the cross section from the origin.
[Given.]

So, A(x) = kx2
[For some constant k of proportion.]

A (4) = 13
[Given.]

k16 = 13 k = 208
[Substitute x = 4 in A(x) = kx2.]

So, A(x) = 208x2

The volume of the solid bounded by x = 2, x = 7 is V = 27 A(x)dx

= 27 208x2dx

= 208(- 1x)27

= 1040 / 14 = 74.29
[Simplify.]

Correct answer : (3)
2.
If $x$ is the distance from the origin to the cross section perpendicular to $x$-axis of a solid, then $x$2 represents the area of the cross section. Find the volume of the solid between the planes $x$ = 1, $x$ = 5 .
 a. 3 cubic units b. 124 cubic units c. $\frac{124}{3}$cubic units d. $\frac{3}{124}$cubic units

#### Solution:

Area of cross section perpendicular to x-axis of the solid = A(x) = x2 where x is the distance of the cross section from the origin.

The volume of the solid bounded by x = 1, x = 5 is

V = 15 A(x)dx

= 15 x2dx

= 1 / 3(x3)15

= 124 / 3 cubic units
[Simplify.]

Correct answer : (3)
3.
A($x$) = $x$2 + 6$x$ represents the area of cross section perpendicular to $x$-axis of a solid when $x$ represents the distance of cross section from the origin. What is the volume of the solid bounded by $x$ = 3, $x$ = 6?
 a. 81 cubic units b. 144 cubic units c. 63 cubic units d. 198 cubic units

#### Solution:

Area of cross section perpendicular to x-axis of the solid = A(x) = x2 + 6x where x is the distance of the cross section from the origin.

The volume of the solid bounded by x = 3, x = 6

= V = 36 A(x)dx

= 36 (x2 + 6x)dx
[Substitute A(x) = x2 + 6x.]

= 1 / 3(x3)36 + 3(x2)36

= (1 / 3)(216 - 27) + 3(36 - 9)

= 144 cubic units
[Simplify.]

Correct answer : (2)
4.
A square based pyramid of height 8 cm is resting on $x$-axis so that its square cross sections are perpendicular to $x$-axis. The vertex of the pyramid is on the plane $x$ = 1 and the base of it is on the plane $x$ = 5. If $x$ is the distance from origin to the cross section, then the area of the cross section is 12$x$2. Find the volume of the pyramid.
 a. 504 b. 496 c. 372 d. 500

#### Solution:

Area of cross section perpendicular to x-axis of the pyramid = A(x) = 12x2 where x is the distance of the cross section from the origin.

The volume of the pyramid bounded by x = 1, x = 5 is V = 1 5 A(x)dx

= 1 5(12x2) dx
[Substitute A(x) = 12x2.]

= 12 / 3(x3)1 5

= 4 (125 - 1) = 496

Correct answer : (2)
5.
A solid is lying alongside the interval [0, $\frac{\pi }{4}$] on the $y$-axis. 9sec $y$ tan $y$ is the area of the cross section of the solid perpendicular to the $y$-axis at the point $y$ of [0, $\frac{\pi }{4}$]. What is the volume of the solid?
 a. 9$\sqrt{2}$ - 9 b. - 9 - 9$\sqrt{2}$ c. 9 - 9$\sqrt{2}$ d. 9$\sqrt{2}$ + 9

#### Solution:

The area of cross section perpendicular to the y-axis of the solid is A(y) = 9sec y tan y where y [0, π4]
[Given.]

The volume of the solid between y = 0, y = π4 is V = 0π/4 A(y) dy

= 0π/4 9 sec y tan y dy
[Substitute A(y) = 9 sec y tan y.]

= 9[sec y]0π/4

= 9 [sec π4 - sec 0]

= 92 - 9

Correct answer : (1)
6.
What is the volume of a solid bounded by the planes $x$ = 1, $x$ = 3 whose area of cross section perpendicular to $x$-axis is proportional to ln$x$ where $x$ is the distance of the cross section from the origin, and whose area of cross section is ln 8 square units at $x$ = 2 ?
 a. (9 ln 3) cubic units b. (- 9 ln 3 + 6) cubic units c. (9 ln 3 + 6) cubic units d. (9 ln 3 - 6) cubic units

#### Solution:

The area of cross section perpendicular to x - axis of the solid is proportional to ln x.

A(x) ln x
[Given.]

A(x) = k ln x
[For some constant k of proportion.]

A(2) = ln 8
[Given.]

k ln 2 = 3 ln 2 k = 3
[Substitute x = 2 in A(x) = k ln x.]

So, A(x) = 3 ln x
[Substitute k = 3 in A(x) = k ln x.]

The volume of the solid bounded by x = 1, x = 3 is V = 1 3 A(x)dx

= 1 3 3ln x dx
[Substitute A(x) = 3 ln x.]

= 3((xln x)13 - 13dx)

= 3((x ln x)13 - (x)13)

= 3(3 ln 3 - 2)

= (9 ln 3 - 6) cubic units
[Simplify.]

Correct answer : (4)
7.
The volume of a solid bounded by the planes $x$ = $\frac{\pi }{4}$, and $x$ = $\frac{\pi }{2}$, whose area of cross section perpendicular to the $x$-axis at $x$ is 7sin $x$. Find the volume of the solid.
 a. $\frac{7}{\sqrt{2}}$ cubic units b. 2 cubic units c. $\sqrt{2}$ cubic units d. $\frac{7}{2}$ cubic units

#### Solution:

Area of cross section of the solid perpendicular to x - axis = A(x) = 7 sin x where x is the distance of the cross section from the origin.

The volume of the solid between x = π4, x = π2 is

V = π/4π/2 A(x) dx

= π/4π/2 7sin x dx
[Substitute A(x) = 7sin x.]

= 7 (- cos x)π/4π/2

= 72 cubic units.
[Simplify.]

Correct answer : (1)
8.
If $x$ is the distance from the origin to the cross section perpendicular to $x$-axis of a solid, then 12 cos $x$ represents the area of cross section. What is the volume of the solid between the planes $x$ = $\frac{\pi }{3}$ and $x$ = $\frac{\pi }{2}$ ?
 a. 6 - 6$\sqrt{3}$ b. 12 - 6$\sqrt{3}$ c. 12 + 6$\sqrt{3}$ d. 12 - $\sqrt{3}$

#### Solution:

The area of cross section perpendicular to x - axis whose distance from origin is x = A(x) = 12 cos x.

The volume of the solid between the planes x = π3, x = π2 is

V = π/3π/2 A(x) dx

= π/3π/2 12 cos x dx
[Substitute A(x) = 12 cos x.]

= 12(sin x)π/3π/2

= 12(2 -3)2

= 6(2 - 3)

= 12 - 63
[Simplify.]

Correct answer : (2)
9.
A($x$) = 15tan $x$ is the area of cross section perpendicular to the $x$-axis of a solid, where $x$ is the distance of the cross section from the origin. What is the volume of the solid between the planes $x$ = $\frac{\pi }{6}$ to $x$ = $\frac{\pi }{3}$ ?
 a. $\frac{ln2}{2}$ b. $\frac{15\left(ln3\right)}{2}$ c. $\frac{15\left(ln2\right)}{2}$ d. $\frac{ln3}{2}$

#### Solution:

The area of cross section of the solid perpendicular to the x-axis whose distance from the origin is x = A(x) = 15tan x

The volume of the solid between the planes x = π6, x = π3 is V = π/6π/3 A(x)dx

= π/6π/3 15tan x dx
[Substitute A(x) = 15tan x.]

= 15(ln sec x)π/6π/3

= 15(ln sec π3 - ln sec π6)

= 15(1 / 2) ln 3
[Simplify.]

= 15(ln3) / 2

Correct answer : (2)
10.
A($x$) = ($\frac{8}{x}$) is the area of cross section perpendicular to the $x$-axis of a solid where $x$ is the distance of the cross section from the origin. Find the volume of the solid between the planes $x$ = 1, $x$ = 17 .
 a. 17(ln 8) cubic units b. 8(ln 17) cubic units c. ln 17 cubic units d. ln 8 cubic units

#### Solution:

The area of cross section perpendicular to the x-axis of the solid = A(x) = 8x where x is the distance from the origin to the cross section.

The volume of the solid between the planes x = 1, x = 17 is V = 117 A(x) dx

= 117 8x dx
[Substitute A(x) = 8x.]

= 8(ln x)117

= 8(ln 17) cubic units
[Simplify.]

Correct answer : (2)

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