Calculate the volume of the region bounded by the coordinate planes and the surface $z=4-x^2-y^2$ Calculate the volume of the region bounded by the coordinate planes and the surface $z=4-x^2-y^2$
Then I think its $x=0, y=0$ and $z=0$ and the surface $z=4-x^2-y^2$
I think the integral is:
$\int_0^{2}\int_0^{\sqrt{{-x^2}+4}}4-x^2-y^2dydx$
is right?
 A: Your integral approach is correct. An easy way to do it is using polar coordinates. @Rezha Adrian Tanuharja said cylindrical coordinates, but you use them for triple integrals which is not the case (you can also solve this problem with $\displaystyle\iiint_E dV$).
Let $x=rcos(\theta)$, $y=rsin(\theta)$.
Your integration region is the circle in the first quadrant:

Then, we can see that $\theta$ should go from 0 (starting from point $C$) to B, and that angle is $90$ degrees. Then $0\leq\theta\leq \frac{\pi}{2}$. It is also easy to see that the radius of the cirlce goes from $0$ to $2$ (because $2^2=x^2+y^2$).
You should also remember that the jacobian when using polar coordinates is $r$, and we set up the following integral:
$\displaystyle\int_0^{{\pi/}{2}}\int_0^2(4-r^2)rdrd\theta=2\pi$ (this is an easy to solve integral, just a polynomial).
A: Use cylindrical coordinate: $z=4-r^{2}\implies r^{2}=4-z$
$$
\begin{aligned}
V&=\frac{1}{4}\int_{0}^{4}\pi r^{2}dz\\
&=\frac{\pi}{4}\int_{0}^{4}(4-z)dz\\
&=2\pi
\end{aligned}
$$
