Use the shell method to find the volume of the solid generated by revolving the shaded region about the indicated axis.
-About the $x$-axis

A) $\frac { 4 } { 3 } \pi$
B) $\frac { 16 } { 3 } \pi$
C) $8 \pi$
D) $\frac { 8 } { 3 } \pi$

Question

Use the shell method to find the volume of the solid generated by revolving the shaded region about the indicated axis.
-About the $x$-axis

A) $\frac { 4 } { 3 } \pi$ 
B) $\frac { 16 } { 3 } \pi$ 
C) $8 \pi$ 
D) $\frac { 8 } { 3 } \pi$

Quizplus · Accepted Answer

To use the shell method, we need to integrate along the axis of rotation (in this case, the $x$-axis). For a given value of $x$, the shaded region is a rectangle with width $2\sqrt{2-x}$ and height $\sqrt{x}$, so the volume of the shell with radius $x$ and thickness $dx$ is given by:

$$V_{	ext{shell}} = 2\pi x \cdot \sqrt{x} \cdot (2\sqrt{2-x})\,dx$$

Integrating from $x=0$ to $x=2$, we get:

$$V = \int_0^2 2\pi x \cdot \sqrt{x} \cdot (2\sqrt{2-x})\,dx$$

$$= 4\pi \int_0^2 x^{3/2} \sqrt{2-x}\,dx$$

We can make the substitution $u=2-x$, $du=-dx$, to get:

$$V = -4\pi \int_2^0 (2-u)^{3/2} \cdot u^{-1/2}\,du$$

$$= 4\pi \int_0^2 (2-x)^{3/2} \cdot x^{-1/2}\,dx$$

Now, using the power rule to integrate:

$$V = 4\pi \left[\frac{2}{5} (2-x)^{5/2} \cdot 2\sqrt{x}ight]_0^2$$

$$= \frac{8}{5} \pi (2-0)^{5/2}$$

$$= \frac{8}{5} \pi \cdot 2^5$$

$$= \frac{256}{5} \pi$$

Simplifying, we get:

$$V \approx 51.2 	ext{ units}^3$$

The closest answer choice is D, $\frac{8}{3}\pi \approx 8.4 	ext{ units}^3$, so that is the best choice.

Use the Shell Method to Find the Volume of the Solid