I was revising my notes and I came across a question, particularly question 5 as seen below.
The working out says to take the volume of the cylinder away from the volume of the hole.
But shouldn't we just find the volume of the hole since that is technically solid when we rotate x = root y around the y axis?
My understanding is that we would only carry out the method in the working out below if we needed to find the volume of the solid portion of a bowl. But this is not what the question is asking. Am I correct?
In the second problem, you are asked for the volume when the region bounded by the curve $y = x^2$, the $x$-axis, and the lines $x = 0$ and $x = 1$ is rotated around the $y$-axis. That region is shown below.
Your instructor observed that the volume when this region is rotated around the $y$-axis is equal to the volume when the region bounded by the curve $y = x^2$, the line $y = 1$, and the $y$-axis is subtracted from the volume of the cylinder formed by rotating the region bounded by the coordinate axes and the lines $x = 1$ and $y = 1$. The cylinder has radius $1$ and height $1$, so its volume is $$V_{\text{cylinder}} = \pi r^2h = \pi (1)(1) = \pi$$ The region bounded by the curve $y = x^2$, the line $y = 1$, and the $y$-axis is shown below.
Since you are integrating with respect to $y$, $0 \leq x \leq 1$, $y = x^2 \implies x = \sqrt{y}$. When $x = 0$, $y = 0$; when $x = 1$, $y = 1$.
Using the disc method, the volume to be subtracted is \begin{align*} V_{\text{hole}} & = \int_{0}^{1} \pi \sqrt{y}^2~dy\\ & = \int_{0}^{1} \pi y~dy\\ & = \pi\frac{y^2}{2}\bigg\vert_{0}^{1}\\ & = \frac{\pi}{2} - 0\\ & = \frac{\pi}{2} \end{align*} Hence, $$V = V_{\text{cylinder}} - V_{\text{hole}} = \pi - \frac{\pi}{2} = \frac{\pi}{2}$$