Volume of solid above $z=r$ and below $z=\sqrt{8-r^2}$

Question

I need to find the volume of solid above the cone $z=r$ and below the hemisphere $z=\sqrt{8-r^2}.$

Here is what I did:

$$ \begin{align*} V&=\int^{2\pi}_0\int^1_0\left(\sqrt{8-r^2}-r\right)r\,dr\,d\theta\\ &=\int^{2\pi}_0\int^1_0\left(r\sqrt{8-r^2}-r^2\right)\,dr\,d\theta\\ &=\int^{2\pi}_0\left(-\frac{1}{3}(8-r^2)^{3/2}-\frac{r^2}{2}\right)\Bigg|^1_0\, d\theta\\ &=\int^{2\pi}_0\frac{14\sqrt{7}-3}{6}\,d\theta\\ &=\frac{28\sqrt{7}-6}{6}\pi \end{align*} $$

The answer should be $13.88$, which is not what I got.

Where did I go wrong?

Answer 1

The significant problem is that $r=1$ is not the projection of the region onto the $xy$ plane. The projection is given by $\sqrt{8-r^2}=r$ or $r=2$.

Hence,

$$V=\int_{0}^{2\pi} \int_{0}^{2} (\sqrt{8-r^2}-r)rdrd\theta$$

$$=2\pi \int_{0}^{2} \left( r\sqrt{8-r^2}-r^2\right) dr$$

$$=\pi \int_{0}^{2} (2r\sqrt{8-r^2}-2r^2) dr$$

$$=\pi \left(\frac{2}{3}((8-0^2)^{\frac{3}{2}}-(8-2^2)^{\frac{3}{2}})-2\frac{(2)^3}{3} \right)$$

$$=\frac{2\pi}{3} \left(16 \sqrt{2}-16 \right)$$

$$=\frac{32\pi}{3}\left(\sqrt{2}-1\right)$$

Answer 2

The exact value is $\;\dfrac{32\pi}3(\sqrt 2-1)$.

It is the solid of revolution generated by the rotation around the $z$-axis of the domain between the line $z=r$ and the arc of parabola $z^2=8r^2$ (in the (r,z)-plane).

As the line and the arc of parabola intersect at the point $(2,2)$ this volume decomposes as:

1. the cone with vertex the origin height and base radius $2$, the volume thereof being $\dfrac{8\pi}3$.
2. The volume of the solid of revolution generated by the arc of parabola $z^2=8-r^2$ between the point $(2,2)$ and the apex $(0,2\sqrt 2$ of the parabola. There's a formula for this volume: $$\int_2^{2\sqrt 2}\pi r^2\,\mathrm d z.$$