Volumes of Revolution — AP Calculus BC practice questions

Question 1

The volume of the solid generated by rotating the region between y = x and y = x^2 on [0, 1] about the x-axis equals:

Accepted Answer

V = π ∫_0^1 (x^2 - x^4) dx = π [(x^3)/(3) - (x^5)/(5)]_0^1 = π ((1)/(3) - (1)/(5)) = (2π)/(15).

Answer

π/30

Answer

π/15

Answer

2π/15

Answer

π/3

Question 2

Rotating y = 2x + 1 on [0, 2] about the x-axis generates a frustum (truncated cone) of volume:

Accepted Answer

V = π ∫_0^2 (4x^2 + 4x + 1) dx = π [(4x^3)/(3) + 2x^2 + x]_0^2 = π ((32)/(3) + 8 + 2) = π · (32 + 30)/(3) = (62π)/(3).

Answer

5π

Answer

14π

Answer

(62π)/(3)

Answer

(62)/(3)

Question 3

Rotating f(x) = √(R^2 - x^2) on [-R, R] about the x-axis generates a:

Accepted Answer

The graph is a semicircle; rotating about the x-axis sweeps out a sphere of radius R. (Volume = (4)/(3)π R^3.)

Answer

cone

Answer

cylinder

Answer

sphere of radius R

Answer

ellipsoid

Question 4

The volume generated by rotating f(x) = √(x) on [0, 4] about the x-axis is:

Accepted Answer

V = π ∫_0^4 (√(x))^2 dx = π ∫_0^4 x dx = π · (16)/(2) = 8π.

Answer

4π

Answer

8π

Answer

12π

Answer

16π

Question 5

The volume generated by rotating f(x) = cos x on [0, π/2] about the x-axis is closest to:

Accepted Answer

V = π ∫_0^π/2 cos^2 x dx = π ∫_0^π/2 (1 + cos 2x)/(2) dx = (π)/(2) [x + (sin 2x)/(2)]_0^π/2 = (π)/(2) · (π)/(2) = (π^2)/(4).

Answer

(π)/(2)

Answer

(π^2)/(4)

Answer

(π^2)/(2)

Answer

π

Question 6

The volume generated by rotating f(x) = x^2 on [0, 1] about the x-axis is:

Accepted Answer

V = π ∫_0^1 x^4 dx = π · (1)/(5) = (π)/(5).

Answer

π

Answer

(π)/(3)

Answer

(π)/(5)

Answer

(π)/(7)

Question 7

The volume generated by rotating f(x) = e^x on [0, 1] about the x-axis is:

Accepted Answer

V = π ∫_0^1 (e^x)^2 dx = π ∫_0^1 e^2x dx = π · dfrace^2x2|_0^1 = (π)/(2)(e^2 - 1).

Answer

π(e - 1)

Answer

π e

Answer

(π)/(2)(e^2 - 1)

Answer

π e^2

Question 8

The region bounded by the graphs of f(x)=x and g(x)=x^2, for x∈[0,1], is rotated about the Ox axis. The volume of the resulting solid is:

Accepted Answer

On [0,1], f(x)=x lies above g(x)=x^2 (the outer and inner radii of the washer, respectively): V = π∫_0^1(x^2-x^4) dx = π[(x^3)/(3)-(x^5)/(5)]_0^1 = π(frac13-frac15)=π·(2)/(15)=(2π)/(15). Check: the option (π)/(3) appears if you drop the term x^4 (inner radius) entirely ✓.

Answer

(π)/(3)

Answer

(2π)/(15)

Answer

(π)/(5)

Answer

(2π)/(5)

Question 9

The volume generated by rotating f(x) = x on [0, 3] about the x-axis is:

Accepted Answer

V = π ∫_0^3 x^2 dx = π · (27)/(3) = 9π. (Equivalently, V_cone = (1)/(3)π r^2 h = (1)/(3)π · 9 · 3 = 9π.)

Answer

3π

Answer

6π

Answer

9π

Answer

27π

Question 10

The volume generated by rotating f(x) = 2 on [0, 5] about the x-axis is:

Accepted Answer

Cylinder of radius 2 and height 5: V = π · 4 · 5 = 20π.

Answer

5π

Answer

10π

Answer

20π

Answer

40π

Volumes of Revolution practice questions — AP Calculus BC

Try 3 problems from this chapter

Problems & worked solutions

Problem 1 — Volumes of Revolution

Problem 2 — Volumes of Revolution

Problem 3 — Volumes of Revolution

Problem 4 — Volumes of Revolution

Problem 5 — Volumes of Revolution

Problem 6 — Volumes of Revolution

Problem 7 — Volumes of Revolution

Problem 8 — Volumes of Revolution

Problem 9 — Volumes of Revolution

Problem 10 — Volumes of Revolution