Using Spherical coordinates find the volume:

Inside the surfaces $z=x^2+y^2$ and $z=\sqrt{2-x^2-y^2}$

I integrated over the ranges:

$0 \leq \theta \leq 2\pi$

$ 0 \leq \phi \leq \frac{\pi}{2}$

$0 \leq r \leq \sqrt{2}$

I get $\frac{\pi}{2}(4\sqrt{2} -4).$

There answer is the same except a $-\frac{7}{2}$ instead of the 4 at the end. Obviously I'm missing a 1\2 but I seems, I can not find it.

Solution 1:

When $z=x^2+y^2=r^2$ and $z=\sqrt{2-x^2-y^2}=\sqrt{2-r^2}$ intersect each other, you will have $$r=1$$ This means that $z=1$ and $\tan(\phi)=1$. So the range of changing for $\phi$ will be $\pi/4\leq\phi\leq\pi$. While typing, I saw @Mhenni did it completely, so I am ending. :-)

Solution 2:

This problem is actually better suited for cylindrical coordinates:

$$\begin{align}\int_{0}^{2\pi}\int_0^1\int_{r^2}^{\sqrt{2-r^2}}rdzdrd\theta&=2\pi\int_0^1(r\sqrt{2-r^2}-r^3)dr\\&=2\pi(-\frac{1}{3}(2-r^2)^{3/2}-\frac{r^4}{4})\mid_{0}^1\\&=2\pi(-\frac{1}{3}-\frac{1}{4}+\frac{2\sqrt{2}}{3})\\&=\frac{\pi}{3}(4\sqrt2-\frac{7}{2})\end{align}$$

The problem with spherical coordinates here is that the radius is a (piecewise) function of the azimuthal angle, which makes the integration a bit more difficult.