For the region in the fourth quadrant bounded by $y=x^2-16$, the $x$-axis, and $y$-axis, determine which of the following is greater: the volume of the solid generated when the region is revolved about the $x$-axis or about the $y$-axis.
The first answer I figured it out which was $\frac{8192}{15} \pi$. However, now it is requesting when the region is revolved about the $y$-axis. Can someone help be with this?
Just evaluate the following:
$$\pi \int_{-16}^0 x^2 \, dy = \pi \int_{-16}^0 y + 16 \, dy$$
If you know the formula for the $x$-axis, you can perform similar trick, we just have to switch the role of $x$ and $y$.
Edit: \begin{align}\pi \int_{-16}^0 x^2 \, dy &= \pi \int_{-16}^0 y + 16 \, dy \\ &= \pi \left[\frac{y^2}2 +16y\right]_{-16}^{0}\\ &=\pi\left[\frac{0^2-(-16)^2}2 + 16(0-(-16)) \right]\\ &=\pi\frac{16^2}2\\ &=2^7\pi \\ &=128\pi\end{align}