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:
$\int \int \int _{K} {\sqrt {4-r^2}rdzdrd\theta }$
 A: This is much easier to solve in cylindrical coordinates. $$x=r\cos\theta\\y=r\sin\theta\\z=h$$
Then the limits for $r$ are $0$ and $1$, the limits for $\theta$ are from $-\frac\pi4$ to $\frac{3\pi}4$, and the limits for $h$ are $0$ and $4-r^2$.
With these, $$V=\int_{-\frac\pi4}^{\frac{3\pi}4}d\theta\int_0^1dr\cdot r\int_0^{\sqrt{4-r^2}}dh$$
Note see comment below. Since $y>0$, the lower limit for $\theta$ is $0$, not $-\pi/4$
A: 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 -
$\displaystyle \int_{0}^{3\pi/4} \int_{0}^{1}  \int_{0}^{\sqrt{4-r^2}} r \, dz \, dr \, d\theta$
A: Using spherical coordinates, you would have to split up $K$ into two regions,
$$K_1=\left\{(r,\theta,\phi)\mid 0\le r\le2,0\le\theta\le\frac{3\pi}4,0\le\phi\le\frac\pi6\right\}$$
$$K_2=\left\{(r,\theta,\phi)\mid0\le r\le\sqrt{\csc\phi},0\le\theta\le\frac{3\pi}4,\frac\pi6\le\phi\le\frac\pi2\right\}$$
(where $x=r\cos\theta\sin\phi$, $y=r\sin\theta\sin\phi$, and $z=r\cos\phi$). The upper limit on $\phi$ for $K_1$ and lower limit for $K_2$ come from the intersection of the cylinder $x^2+y^2=1$ and the sphere $z=\sqrt{4-x^2-y^2}$. On the sphere, $r=2$, so we have
$$2\cos\phi=\sqrt3\implies\phi=\cos^{-1}\left(\frac{\sqrt3}2\right)=\frac\pi6$$
The upper limit for $r$ in $K_2$ is obtained by converting the equation of the cylinder $x^2+y^2=1$ into spherical coordinates:
$$(r\cos\theta\sin\phi)^2+(r\sin\theta\sin\phi)^2=r^2\sin^2\phi=1\implies r=|\csc\phi|=\csc\phi$$
Then the volume is
$$\int_0^{\frac\pi6}\int_0^{\frac{3\pi}4}\int_0^2r^2\sin\phi\,\mathrm dr\,\mathrm d\theta\,\mathrm d\phi+\int_{\frac\pi6}^{\frac\pi2}\int_0^{\frac{3\pi}4}\int_0^{\csc\phi}r^2\sin\phi\,\mathrm dr\,\mathrm d\theta\,\mathrm d\phi$$
The first integral is trivial. For the second, integrating with respect to $r$ yields
$$\int_{\frac\pi6}^{\frac\pi2}\int_0^{\frac{3\pi}4}\int_0^{\csc\phi}r^2\sin\phi\,\mathrm dr\,\mathrm d\theta\,\mathrm d\phi=\frac13\int_{\frac\pi6}^{\frac\pi2}\int_0^{\frac{3\pi}4}\csc^2\phi\,\mathrm d\theta\,\mathrm d\phi$$
and observing that $\csc^2\phi=\frac{\mathrm d}{\mathrm d\phi}(-\cot\phi)$, it turns out the second integral is, too.
