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

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

  2. Using variables change theorem, calculate the next integral using polar coordinates:

$$ \iint\limits_D \frac{1}{1+x^2+y^2}dxdy$$ where $D=\{(x,y)\in \mathbb{R}^2|y\in [0, 1] \text{ and } 0\leq x \leq \sqrt{1-y^2}\}$

$$x=r\cos\theta\\ y=r\sin\theta $$

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Some of your formatting went astray. Can you please correct? Regards –  Amzoti Jan 16 '13 at 19:41
    
I'm sorry, I'd try to edit, but I cannot understand this. –  Ron Gordon Jan 16 '13 at 19:41
    
i tried to edit ... check it out. if anything wen't amiss, notify!! –  Santosh Linkha Jan 16 '13 at 19:47
    
Thanks for editing. Regards! –  user58516 Jan 16 '13 at 19:55
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In general, it is good practice to state from where these integrals arise and to state your thoughts on the matter. –  JavaMan Jan 16 '13 at 20:39

5 Answers 5

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 $$\begin{align} \iiint\limits_D y^2 \; dx \; dy \; dz &= {1 \over 2} \iiint\limits_B y^2 \; dx \; dy \; dz = {1 \over 2} \iiint\limits_B x^2 \; dx \; dy \; dz = {1 \over 2} \iiint\limits_B z^2 \; dx \; dy \; dz\cr &= {1 \over 6} \iiint\limits_B(x^2+y^2+z^2) \; dx \; dy \; dz\cr &= {1 \over 6} \int_0^1\iint\limits_\Sigma \rho^2 \; dS \; d\rho = {1 \over 6} \int_0^1\iint\limits_\Sigma \; dS \;\rho^2\; d\rho = {1 \over 6} \int_0^1 4\pi\rho^2 \;\rho^2\; d\rho \cr &= {1\over6} {4\pi \over 5} = {2 \pi \over 15}\,. \end{align} $$

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+1 This is a very nice approach. –  DonAntonio Jan 17 '13 at 2:37

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.

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1  
Still checking ...hold on. –  DonAntonio Jan 16 '13 at 20:31
    
I get $\frac{2\pi}{15}$ on wolfram: link using @experimentX 's formulation –  Rustyn Jan 16 '13 at 20:41
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I confirmed your computation on MATLAB, (faster) \begin{verbatim} >> syms r >> syms theta >> syms z >> f=r^3*sin(theta).^2 >> int(int(int(f,z,0,sqrt(1-r^2)), theta,0,pi),r,0,1) ans = pi/15 \end{verbatim} –  Rustyn Jan 16 '13 at 21:02
    
What J is in this case? –  user58516 Jan 16 '13 at 21:05
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I think $-\sqrt{1-r^2}\le z \le \sqrt{1-r^2}$, which would explain the missing factor of 2. –  Michael E2 Jan 16 '13 at 21:31

I'll simplify question $2$ for you:
$$x=r\cos\theta\\ y=r\sin\theta$$ and $$ D=\{(x,y)\in \mathbb{R}^2|y\in [0, 1] \text{ and } 0\leq x \leq \sqrt{1-y^2}\} $$ $\Rightarrow$
$$ \iint\limits_D \frac{1}{1+x^2+y^2}dxdy = \int_{0}^{\tfrac{\pi}{2}}\int_{0}^{1} \frac{r}{1+r^2} drd\theta = ... = \frac{\pi}{4}\log(2) $$

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this is the first time i ever wrote so much!! +1 –  Santosh Linkha Jan 16 '13 at 20:25
    
looks like you forgot to edit the final value .. since limit has been changed. –  Santosh Linkha Jan 16 '13 at 20:28
    
@experimentX Thanks, (whoops) –  Rustyn Jan 16 '13 at 20:30

$$\iint\limits_D \frac{1}{1+x^2+y^2}dxdy = \int_{0}^{\pi/2}\int_{0}^{1} \frac{1}{1+r^2}rdrd\theta$$

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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 $$ \int_0^1\int_0^{\pi \over 2} \frac{1}{1 + (r \cos \theta)^2 + (r\sin \theta)^2} r d\theta dr $$

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The limits of integration of the outer integral should be from $0$ to $1$. –  Mhenni Benghorbal Jan 16 '13 at 20:22
    
@MhenniBenghorbal thank you for correction!! –  Santosh Linkha Jan 16 '13 at 20:23
    
@experimentX whoops I had $\pi/4$ on my limits of integration (accident), +1 for in-depth explanation. Rustyn Yazdanpour would like to appeal to the OP to accept experimentX's answer in place of his –  Rustyn Jan 16 '13 at 20:25
    
@experimentX: You are welcome. –  Mhenni Benghorbal Jan 16 '13 at 20:29
    
@experimentX yours and Don Antonio's answers do not concur: \begin{verbatim} >> syms phi >> f1=(r*sin(theta)*sin(phi) )^2*(r^2*sin(theta)) f1 = r^4*sin(phi)^2*sin(theta)^3 >> int(int(int(f1,phi,0,pi),theta,0,pi),r,0,1) ans = (2*pi)/15 \end{verbatim} I'm going to eat some lunch. –  Rustyn Jan 16 '13 at 21:15

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