Region between y = 2 - x and y = 0 on [0, 2] revolved about the x-axis:

\tfrac{8\pi}{3} V = \pi\int0^2 (2-x)^2\,dx = \pi[-( 2 - x)^3/3]0^2 = \pi(0 + 8/3) = 8\pi/3.

Volumes by Disk and Washer Methods

AP Calculus AB· difficulty 3/5

y = 2 - x

Explanation

V = \pi\int_0^2 (2-x)^2\,dx = \pi[-( 2 - x)^3/3]_0^2 = \pi(0 + 8/3) = 8\pi/3

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