Below is a problem I did. The book gets $\frac{16\pi}{15}$.This number seems to large to me. I am hoping that somebody can confirm that I got it right or tell me where I went wrong.
Problem:
Find the volume of the solid obtained by rotating the region bounded by the given lines and curves about the x-axis. $$ x = 1 - y^2 $$ $$ x = 0 $$
Answer:
First, I plot the function. A plot of the function
If you look at the plot, a rough estimate of the answer would be to treat it as a cylinder. Looking at the graph, a rough estimate of the radius of the cylinder would be $\frac{3}{4}$ and its height is $1$. Therefore, a rough estimate of the answer would be $\pi ( \frac{3}{4} )^2 $ which is $\frac{9 \pi }{16}$.
Let $V$ be the volume we seek. \begin{align*} V &= \int_0^1 \pi y^2 \,\, dx \\ % x - 1 &= -y^2 \\ y^2 &= 1 - x \\ V &= \int_0^1 \pi ( 1-x ) \,\, dx \\ V &= \pi \int_0^1 ( 1-x ) \,\, dx \\ \int_0^1 ( 1-x ) \,\, dx &= x - \frac{x^2}{2} \Big|_{0}^{1} = \frac{1}{2} \\ V &= \frac{\pi}{2} \end{align*}
After double checking the book, I found that it says about the y-axis, not the x-axis. I am still hoping that somebody here can confirm that my work is right if you go about the x-axis.