Two Multiple Integrals Could anyone solve this two integrals? I would really appreciate. Thank you!


*

*$$\iiint\limits_D y^2 dx dy dz$$
where $D=\{(x,y,z)|y\geq 0 \text{ and } x^2+y^2+z^2 \leq 1\}$

*Using variables change theorem, calculate the next integral using polar coordinates:
$$ \iint\limits_D \frac{1}{1+x^2+y^2}dxdy$$ 
where $D=\{(x,y)\in \mathbb{R}^2|y\in [0, 1] \text{ and } 0\leq x \leq \sqrt{1-y^2}\}$
$$x=r\cos\theta\\
y=r\sin\theta
$$
 A: For example (1): cylindrical coordinates
$$x=r\cos\theta\;\;,\;\;y=r\sin\theta\;\;,\;\;z=z\;\;,\;\;J=r\;\;,\;\;0\leq \theta\leq \pi\Longrightarrow$$
$$\int\limits_0^1dr\int\limits_0^{\pi}d\theta\int\limits_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}}r^3\sin^2\theta\, dz=2\int\limits_0^1r^3\sqrt{1-r^2}\,dr\int\limits_0^\pi \sin^2\theta\,d\theta=$$
$$=\frac{4}{15}\int\limits_0^\pi \sin^2\theta\,d\theta=\frac{4}{15}\left.\left(\frac{\theta-\sin\theta\cos\theta}{2}\right)\right|_0^\pi=\frac{2\pi}{15}$$
Disclaimer: Check carefully the above.
Added: Thanks to Michael E2 for the comment below. The correction's been done.
A: I'll simplify question $2$ for you:  
$$x=r\cos\theta\\
y=r\sin\theta$$
and 
$$
D=\{(x,y)\in \mathbb{R}^2|y\in [0, 1] \text{ and } 0\leq x \leq \sqrt{1-y^2}\}
$$
$\Rightarrow$ 
$$
\iint\limits_D \frac{1}{1+x^2+y^2}dxdy = \int_{0}^{\tfrac{\pi}{2}}\int_{0}^{1} \frac{r}{1+r^2} drd\theta = ... = \frac{\pi}{4}\log(2)
$$
A: For the first integral, we are evaluating $y^2$ along the hemisphere $ y\geq 0 \text{ and } x^2+y^2+z^2 \leq 1$ of radius 1. Without $y^2$, it should give the volume of hemisphere. We can change into spherical coordinate system. The only constraint is $y \ge 0$ so the limit would be as usual except for $0\le \theta \le \pi $, we have
$$ \int_0^1 \int_0^\pi \int_0^\pi (r \sin \theta \sin \phi )^2\; ( r^2 \sin \phi) \; d\phi d\theta dr $$
For the second integral, note that $ 0\leq x \leq \sqrt{1-y^2} \implies x^2+y^2 \le 1$ which is a circle of radius $1$ also $x\ge 0$ and $y \in [0, 1] $0 implies we are evaluating it in the first quadrant of circle. Change into polar coordinate system. The only constraint is in $\theta$ which goes from $0$ to $\pi \over 2$. So the integral is
$$ \int_0^1\int_0^{\pi \over 2} \frac{1}{1 + (r \cos \theta)^2 + (r\sin \theta)^2} r d\theta dr $$
A: $$\iint\limits_D \frac{1}{1+x^2+y^2}dxdy = \int_{0}^{\pi/2}\int_{0}^{1} \frac{1}{1+r^2}rdrd\theta$$
A: A different approach to (1): Letting $B$ be the unit ball and $\Sigma$ be the surface of a ball of radius $\rho$, then by symmetry
$$\begin{align}
\iiint\limits_D y^2 \; dx \; dy \; dz
 &= {1 \over 2} \iiint\limits_B y^2 \; dx \; dy \; dz
 = {1 \over 2} \iiint\limits_B x^2 \; dx \; dy \; dz
 = {1 \over 2} \iiint\limits_B z^2 \; dx \; dy \; dz\cr
 &= {1 \over 6} \iiint\limits_B(x^2+y^2+z^2) \; dx \; dy \; dz\cr
 &= {1 \over 6} \int_0^1\iint\limits_\Sigma \rho^2 \; dS \; d\rho
  = {1 \over 6} \int_0^1\iint\limits_\Sigma  \; dS \;\rho^2\; d\rho
  = {1 \over 6} \int_0^1 4\pi\rho^2 \;\rho^2\; d\rho \cr
 &= {1\over6} {4\pi \over 5} = {2 \pi \over 15}\,.
\end{align}
$$
