Find the volume of the solid generated by revolving the region about the given axis. The region is bounded by $y = 6\sqrt{x}$, $y = 6$, and $x = 0$ about the line $x = 1$.
I need help checking my answer. I got $\displaystyle\frac{24\pi}{5}$ but the answer key I have says the answer is $\displaystyle\frac{14\pi}{5}$. The answer key only shows the answer but no work. Can someone show me the steps for how to solve this problem correctly? i used the shell and washer method and arrived at the same answer.
Let me start off by drawing the boundaries on a Cartesian plane:
To find the volume we rotate this area around the line $x = 3$, which is indeed a well posed problem. Let's take a small sliver of the area of with thickness $dx$. We choose our slice on the $x$-axis for simplicity, but this can be done with a slice of thickness $dy$ on the $y$-axis as well. Now we have the following diagram:
The volume contributed by this slice can be calculated by noting that this slice generates a cylindrical shell after rotation. That is,
$$dV = 2 \pi \left( \mathrm{radius} \right) \left( \mathrm{strip \, area} \right) = 2 \pi \left( 1 - x \right) \left[ \left(6 - 6 \sqrt{x} \right) dx \right]$$
We can now integrate this expression to obtain the full volume:
$$V = \int_{0}^{1} \left( 1 - x \right) \left[ \left(6 - 6 \sqrt{x} \right) dx \right] = 12 \pi \int_{0}^{1} \left(1 - x - x^{\frac{1}{2}} + x^{\frac{3}{2}} \right) dx = \frac{14 \pi}{5}$$
Hope that helps!