Consider the cone $z^2=x^2+y^2$ between $z=0$ and $z=1$. Find the volume of the region above this cone and inside the sphere of radius $\sqrt2$ centered at the origin that encloses the cone.
The straightforward approach to this problem would have been a triple integral in spherical coordinates, but for practice I tried using the cone as a surface and using a surface integral to find the volume between the cone and the sphere. I first parameterized the surface using cylindrical coordinates $x=r\cos\theta$, $y=r\sin\theta$, $z=r$ (since $z^2=x^2+y^2$), and then found the intersection points using the equations $z=r$ and $r^2+z^2=2$: $$2z^2=2\Rightarrow z=1\Rightarrow r=1$$ Now that the bounds of integration have been found I used the function $z=\sqrt{2-r^2}$ as the function to integrate and set up my surface integral as: $$\iint_S \sqrt{2-r^2}\Vert\vec{r_r}\times\vec{r_\theta}\Vert\,\text{d}S$$ I then computed the cross product, for $\vec{r}(r,\theta)=\left<r\cos\theta,r\sin\theta,r\right>$: $$\begin{vmatrix} \hat{\imath}&\hat{\jmath}&\hat{k}\\ \cos\theta&\sin\theta&1\\ -r\sin\theta&r\cos\theta&0 \end{vmatrix}=\hat{\imath}(-r\cos\theta)-\hat{\jmath}(-r\sin\theta)+\hat{k}(r\cos^2\theta+r\sin^2\theta)$$ So the surface integral becomes $$\int_0^{2\pi}\int_0^1\sqrt{\left(2-r^2\right)\left(r^2\cos^2\theta+r^2\sin^2\theta+r^2\right)}\,\text dr\,\text d\theta$$ $$=\int_0^{2\pi}\int_0^1\sqrt{\left(2-r^2\right)r^2\left(\cos^2\theta+\sin^2\theta+1\right)}\,\text dr\,\text d\theta$$ $$=\int_0^{2\pi}\int_0^1\sqrt{2r^2(2-r^2)}\,\text dr\text d\theta$$ $$=2\sqrt2\pi\int_0^1r\sqrt{2-r^2}\,\text dr$$ Using a $u$-substitution of $u=2-r^2$: $$\sqrt2\pi\int_1^2 \sqrt{u}\,\text du=\sqrt2\pi\left.\left(\frac{2u\sqrt u}{3}\right)\right|_1^2=\sqrt2\pi\left(\frac{4\sqrt2-2}{3}\right)$$ $$=\frac{2\pi}{3}\left(4-\sqrt2\right)$$ However, the answer given in the answer key says that the volume is $\frac{4\pi}{3}(\sqrt2-1)$, so I would like to know why my answer is incorrect, whether that be a computational mistake or faulty reasoning as to why a surface integral would work here to find the volume.
EDIT: Here is the posted solution:
