AP Calculus AB · Topic 8.6

Volumes by Disk and Washer Methods Practice

Part of Applications of Integration.(CHA-4.F)

Practice questions

25

Want a predicted score for the whole AP CAL exam? Take the 20-question diagnostic and Lumi will plan the rest.

Sample questions

5 of 25 — sign in to practice the rest with adaptive difficulty and mastery tracking.

  1. Sample 1difficulty 2/5

    For y=cy = c (constant) revolved on [a,b][a, b] about xx-axis: Volume is

    • A

      πc(ba)\pi c (b - a)

    • B

      c(ba)c (b - a)

    • C

      πc2(ba)\pi c^2 (b - a)

      check_circle
    • D

      2πc(ba)2\pi c (b - a)

    Why

    Cylinder volume: πr2h=πc2(ba)\pi r^2 h = \pi c^2 (b - a).

  2. Sample 2difficulty 2/5

    r=f(x)

    For revolution about the xx-axis, the radius of the disk at location xx is

    • A

      f(x)f(x)

      check_circle
    • B

      x2x^2

    • C

      xx

    • D

      f(x)f'(x)

    Why

    Distance from axis to curve at xx is f(x)f(x).

  3. Sample 3difficulty 3/5

    Cone of radius RR, height HH via y=(R/H)xy = (R/H) x, 0xH0 \le x \le H, revolved:

    • A

      23πR2H\tfrac{2}{3}\pi R^2 H

    • B

      13πR2H\tfrac{1}{3}\pi R^2 H

      check_circle
    • C

      πR2H\pi R^2 H

    • D

      12πR2H\tfrac{1}{2}\pi R^2 H

    Why

    Standard cone volume formula confirmed by integration.

  4. Sample 4difficulty 3/5

    The disk method computes a volume of revolution around the xx-axis as

    • A

      ab2πf(x)dx\int_a^b 2\pi f(x)\,dx

    • B

      abf(x)dx\int_a^b f(x)\,dx

    • C

      abπ[f(x)]2dx\int_a^b \pi [f(x)]^2\,dx

      check_circle
    • D

      ab2πxf(x)dx\int_a^b 2\pi x\,f(x)\,dx

    Why

    Each disk has radius f(x)f(x) and thickness dxdx → volume π[f(x)]2dx\pi[f(x)]^2\,dx.

  5. Sample 5difficulty 3/5

    Rotate x=y2x = y^2, 0y20 \le y \le 2, about the yy-axis. Volume by disks (in yy):

    • A

      π02y2dy=8π3\pi\int_0^2 y^2\,dy = \dfrac{8\pi}{3}

    • B

      02y4dy\int_0^2 y^4\,dy

    • C

      2π02y2dy2\pi\int_0^2 y^2\,dy

    • D

      π02y4dy=32π5\pi\int_0^2 y^4\,dy = \dfrac{32\pi}{5}

      check_circle

    Why

    Disks of radius x=y2x = y^2, thickness dydy: πy4dy\pi y^4\,dy.