Volume of solid revolution 2

Question

Let R be the be the limited area in the first quadrant that's above the curve $ y = x^2$ and below the curve $y = 8 - x$. Determine the volume when it's rotated around the y-axis.

The problem for me here is that it's rotated around the y-axis. Otherwise I could just have used the function $\int_0^2 \pi (8-x)^2 - \pi (x^2)^2$ but that doesn't work here.

How do I tackle problems like these?

Answer 1

The wanted solid of revolution is given by the union of a cone and a parabolic cap:

Since $\sum_{r=1}^{n}r = \frac{n^2}{2}+O(n)$ and $\sum_{r=1}^{n}r^2 = \frac{n^3}{3}+O(n^2)$ we have that the volume of the cone is $\frac{1}{3}\pi R^2 h_1$ and the volume of the parabolic cap is $\frac{1}{2}\pi R^2 h_2$. Since the graphs of $y=8-x$ and $y=x^2$ meet at $\left(\frac{-1+\sqrt{33}}{2},\frac{17-\sqrt{33}}{2}\right)$ in the first quadrant, we have $R=h_1=\frac{-1+\sqrt{33}}{2}$ and $h_2=\frac{17-\sqrt{33}}{2}$, so the wanted volume is $$ V = \frac{\pi}{12}\left(33\sqrt{33}-433\right).$$

Answer 2

Hint: for $x,y\geq 0$, you may invert both functions to find $$ x=y-8\\ x=\sqrt{y} $$ and proceed with your usual methods.