MATH 2414 - Volumes of Solids with Known Cross Sections Exercises
1. Find the volume of the solid that has cross-sectional area A(x) = x + 2 , for –1 ≤ x ≤ 3.
Answer: 12
2. Find the volume of the solid that has cross-sectional area A(x) = π(4 − x) 2 , for 0 ≤ x ≤ 2.
56π
€
Answer:
3
3. Use integration to find the volume of a pyramid€of height 160 ft that has a square base measuring
300 ft on each side.
Answer:
4,800,000 ft3
€
⎛ x ⎞
4. A cookie jar has circular cross sections of radius 4 + sin⎜ ⎟ inches for 0 ≤ x ≤ 2π. Sketch a
⎝ 2 ⎠
drawing of the jar and calculate its volume.
(
)
Answer: 33π 2 + 32π in3
€
5. Find the volume of the solid whose base is an isosceles right triangle with legs that are each 4 ft
long and every cross section perpendicular to one of the legs is a semicircle.
€
8π 2 3
Answer:
ft
3
6. Find the volume of the solid whose base is a circle with a radius of 1 ft and whose cross sections
perpendicular to a fixed diameter of the base are squares.
16 3
€
Answer:
ft
3
7. Find the volume of the solid whose base is an equilateral triangle each side of which has a length
of 10 inches and whose cross sections perpendicular to a given altitude of the base are squares.
€
500 3 3
Answer:
in
3
8. The base of a solid is bounded by y = x 3 , y = 0 , and x = 1. Find the volume of the solid for each
of the following cross sections taken perpendicular to the y-axis: (a) squares, (b) semicircles,
(c)
€ equilateral triangles, and (d) semiellipses whose heights are twice the lengths of their bases.
3 2
€
Hint for Part (c): The area€of an equilateral
triangle
is
s , where s is the length of a side.
€
4
Hint for Part (d): The area of an ellipse is πab , where 2a and 2b are the lengths of the axes.
1
π
π
3
Answer: (a)
(b)
(c)
(d)
10
80
20 €
40
€
9. The base of a solid is bounded by y = 3 x , x = 0 , and y = 1. Find the volume of the solid for
each€of the following
taken perpendicular to the x-axis: (a) squares, (b) semicircles,
€cross sections
€
€
(c) equilateral triangles, and (d) semiellipses whose heights are twice the lengths of their bases.
3 2
€
Hint for Part (c): The area of
triangle
s , where s is the length of a side.
€ is
€ an equilateral
4
Hint for Part (d): The area of an ellipse is πab , where 2a and 2b are the lengths of the axes.
1
π
π
3
Answer: (a)
(b)
(c)
(d)
10
80
20 €
40
€
€
€
€
€
Solutions to Volumes of Solids with Known Cross Sections Exercises
1.
3
∫−1( x + 2)dx
2.
2
∫0 π (4 − x)2dx
3. Refer to the drawing at the right, which represent the
horizontal midsection of the pyramid.
€
150−0
15
15
m=
= − so y = − x +150 . The length of a side of the
€
0−160
16
16
(
)
2
square is 2y, so −15 x + 300 . Thus A( x) = −15 x + 300 and
8
€
V=
2
160 −15
€ x + 300 dx .
8
(
∫0
8
)
€
€
4. Because the cross sections of this solid are circles, it is a solid of revolution.
€
2π
∫0
So V =
(
)
2
π 4 + sin x dx .
2
5. Refer to the drawing at the right.
m = 2 2−0 = −1 so y = −x + 2 2 , which is the radius of the
€
0−2 2
semicircle.
1 πr 2
2
(−x + 2 2 )
2
2 21
and V = ∫
π ( −x + 2 2 ) dx .
2
0
Then A( x)
€=
€
=
1π
2
4 2
2
€
€
2 2
€
6. The equation of the circle is x 2 + y 2 = 1 so y = ± 1 − x 2 and the length of a side of the square is
2
1 €
2 1 − x 2 and the area of the square is 4 1 − x 2 . Thus, V = ∫ 4 1 − x 2 dx .
(
€
7. V =
€
∫
2
5 3 2
− x +10 dx
3
0
(
8. (a) V =
€
(c) V =
€
)
)
−1
)
€
€
€
2
1 1 ⎛ 1− y1/3 ⎞
π⎜
dy
0 2 ⎝ 2 ⎟⎠
2
1
1 − y1 / 3 dy
0
(b) V =
∫
2
1 3
1 − y1 / 3 dy
4
(d) V =
∫0 π⎜⎝ 1− y2
∫
∫0
(
)
(
)
€
9. Similar to Exercise 8, but with roles of x and y reversed.
€
(
€
1
⎛
1/3 ⎞
(
⎟ 1 − y
⎠
1/ 3
)dy