Find the volume of the solid generated by revolving the region about the given line.
-The region bounded above by the line $y = 9$, below by the curve $y = 9 - x ^ { 2 }$, and on the right by the line $x = 3$, about the line $y = 9$
A) $\frac { 648 } { 5 } \pi$
B) $9 \pi$
C) $\frac { 243 } { 5 } \pi$
D) $\frac { 567 } { 5 } \pi$

Question

Quizplus · Accepted Answer

To find the volume, we use the shell method. The radius of the shell at a given height $y$ is $9-y$. The height of the shell is $2\sqrt{9-y-x^2}$ (since the distance between the top and bottom curves is $y = 9 - x^2$ and we are revolving around the line $y=9$). The width of the shell is $dx$. Thus, the volume is given by:

$$V=\int_{0}^{3}\int_{9-x^2}^{9}(2\sqrt{9-y-x^2})(9-y)dydx$$

We integrate first with respect to $y$ to get:

$$V=\int_{0}^{3}\left(2\int_{9-x^2}^{9}\sqrt{9-y-x^2}(9-y)dy\right)dx$$

Using the substitution $u=9-y-x^2$, we get:

$$V=\int_{0}^{3}\left(2\int_{0}^{9-x^2}\sqrt{u}u du\right)dx$$

Simplifying and computing the integrals, we obtain:

$$V=\frac{243}{5}\pi$$

Therefore, the answer is (C).

Find the Volume of the Solid Generated by Revolving the Region