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?

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 $$x=rcos(\theta)$$, $$y=rsin(\theta)$$.

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).
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}
• @JohnWaylandBales you mean when I substituted $r^{2}=4-z$? Commented Nov 7, 2020 at 12:59