Volume with spherical polar coordinates Determine the volume between the surface $z=\sqrt{4-x^2-y^2}$ and the area of the xy plane determined by $x^2+y^2\le 1,\ x+y>0,\ y\ge 0$.
I convert to spherical polar coordinates.
$$K=0\le r\le 1,\ 0\le \phi \le \frac{3\pi}{4},\ 0\le \theta \le 2\pi$$
$$\iiint_{K} (\sqrt {4-r^2\sin^2\phi \cos^2\theta-r^2\sin^2\phi \sin^2\theta)}r^2\sin\phi drd\phi d\theta$$
I can't figure out how to take $\int_{K} (\sqrt {4-r^2\sin^2\phi \cos^2\theta-r^2\sin^2\phi \sin^2\theta)}r^2\sin\phi dr$, which makes me think I made a mistake somewhere.
EDIT: Thanks for all the answers.
Now I understand how the limits of $\theta ,r,z$ works.
I don't fully understand where the function "disappear".
$\sqrt {4-x^2-y^2} =\sqrt {4-r^2}$
Why isn't it then:
$\int \int \int _{K} {\sqrt {4-r^2}rdzdrd\theta }$

Area on XY plane is bound by $x^2 + y^2 \leq 1, y \geq 0, x + y \geq 0$
This is a sector of the circle $x^2 + y^2 \leq 1$ bound between positive $X$-axis and line $y = -x$ in the second quadrant. This comes from the fact that $y \geq 0$ so part of the circle in third and fourth quadrant of $XY$ plane is not included. $x + y \geq 0$ is true for quarter of the circle in the first quadrant as both $x$ and $y$ are positive. It is also true for part of the circle in the second quadrant above line $y = -x$ as $|y| \geq |x|$.
Now you are asked to find the volume between this area on XY plane and $z = \sqrt{4-x^2-y^2}$. So it is essentially a cylinder ($\frac{3}{8}$ cross section of a cylinder of radius $1$) cut out of the sphere of radius $2$ above $XY$ plane.
So here is how it will look in cylindrical coordinates -
$\displaystyle \int_{0}^{3\pi/4} \int_{0}^{1}  \int_{0}^{\sqrt{4-r^2}} r \, dz \, dr \, d\theta$
