Suppose $R$ is the region in the first quadrant bounded by $y = 2+x$, $y= x^2$, and $x=0$. I was supposed to find
(a) the volume of the solid generated by revolving around the $y$-axis and
(b) the volume of the solid generated by revolving around $x=3$.
For (a), my answer was exactly $1/2$ of the right answer $(16\pi/3)$, and for (b) my answer was equal to $-2$ times the right answer $(44\pi/3)$.
For (a) I used $\int_0^4(\pi y^2-\pi(y-2)^2)dy.$ I solved the initial equations for $y$ since its around the $y$-axis. And for (b) I used $\int_0^4\left(\pi(\sqrt y-3)^2-\pi(y-5)^2\right)dy.$
Can anybody tell me what I did wrong? Thanks.
a) $\int_0^4\pi ydy-\int_2^4\pi(y-2)^2dy=8\pi-\frac83\pi=\frac{16}3\pi.$
b) $\int_0^2\pi3^2dy+\int_2^4\pi(5-y)^2dy-\int_0^4\pi(3-\sqrt y)^2dy=18\pi+\frac{26}3\pi-12\pi=\frac{44}3\pi.$