A region in the first quadrant is bounded above by and below by for

MATHEMATICS · Single correct · AP · Calculus AB · Applications of Integration · Applications of Integration

Problem

A region in the first quadrant is bounded above by

y=\sqrt{x}

and below by

y=\dfrac{x}{2}

for

0\le x\le4

. The region is revolved about the x-axis. Which of the following integral expressions correctly represents the volume

V

of the solid? (a)

V=\pi\displaystyle\int_{0}^{4}\left[\left(\dfrac{x}{2}\right)^{2}-(\sqrt{x})^{2}\right]dx

(b)

V=2\pi\displaystyle\int_{0}^{4}(\sqrt{x})\left(\dfrac{x}{2}\right)dx

(c)

V=\pi\displaystyle\int_{0}^{4}\left[(\sqrt{x})^{2}-\left(\dfrac{x}{2}\right)^{2}\right]dx

(d)

V=\pi\displaystyle\int_{0}^{2}\left[(\sqrt{x})^{2}-\left(\dfrac{x}{2}\right)^{2}\right]dx