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 March 26th, 2014, 12:40 AM #1 Newbie   Joined: Mar 2014 Posts: 2 Thanks: 0 Triple integral Hi. Now i learning. I solve almost all examples from my book, but thease two remain, there are to hard to solve them. Can someone to show the solution ? Thanks $\int_{d}^{ } \int_{}^{ } ln(1+x^2+y^2)dxdy, d: x^2+y^2=1, x>=0, y>=0$ $\int_{d}^{ } \int_{}^{ } sqrt((x^2)/1+(y^2)/4)), d: x^2+y^2/4=1$
March 26th, 2014, 06:20 AM   #2
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Re: Triple integral

First, these are not "triple" integrals. They are just double integrals.

Quote:
 Originally Posted by Math_user Hi. Now i learning. I solve almost all examples from my book, but thease two remain, there are to hard to solve them. Can someone to show the solution ? Thanks $\int_{d}^{ } \int_{}^{ } ln(1+x^2+y^2)dxdy, d: x^2+y^2=1, x>=0, y>=0$
You are integrating over the unit disk and the integral involves "$x^2+ y^2$" so changing to polar coordinate looks simplest. In polar coordinates, r goes from 0 to 1 while $\theta$ goes from 0 to $\frac{\pi}{4}$. Of course, $dxdy$ becomes $rdrd\theta$. The integral becomes
$\int_{\theta= 0}^{\pi/4}\int_{r= 0}^1 ln(1+ r^2) r dr d\theta= \frac{\pi}{4} \int_0^1 ln(1+ r^2) rdr$
The substitution $u= 1+ r^2$ seems obvious.

Quote:
 $\int_{d}^{ } \int_{}^{ } sqrt((x^2)/1+(y^2)/4)), d: x^2+y^2/4=1$
Here the integration is over the ellipse. The substitution, $x= r cos(\theta)$, $y= 2r sin(\theta)$, a variation on polar coordinates looks like it should work. That way the boundary of the
region is $x^2+ y^2/4= r^2 cos^2(\theta)+ 4r^2 sin^2(\theta)/4= r^2(sin^2(\theta)+ cos^2(\theta)= r^2= 1$, the circle with radius 1. The integral becomes
$\int_{\theta= 0}^{2\pi}\int_{r= 0}^1 \sqrt{r^2cos^2(\theta)+4r^2sin^2(\theta)/4}(\sqrt{2}r dr\theta)= \sqrt{2}\int_{\theta= 0}^{2\pi}\int_{r= 0}^1 r^2 dr d\theta$

 March 26th, 2014, 08:13 AM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,614 Thanks: 2603 Math Focus: Mainly analysis and algebra Re: Triple integral The OP's question states that $x^2 + y^2= 1$ rather than $x^2 + y^2=<= 1$ which suggests that $r^2= 1$ is constant and we thus have a single integal with respect to $\theta$. Similarly for the second example.
 March 26th, 2014, 01:40 PM #4 Newbie   Joined: Mar 2014 Posts: 2 Thanks: 0 Re: Triple integral Thanks, you are awesome. P.S. Yes, i know there are double, just little bit confusing until i try to look in book and to write here.
March 27th, 2014, 06:18 AM   #5
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Re: Triple integral

Quote:
 Originally Posted by v8archie The OP's question states that $x^2 + y^2= 1$ rather than $x^2 + y^2=<= 1$ which suggests that $r^2= 1$ is constant and we thus have a single integal with respect to $\theta$. Similarly for the second example.
I suspect that the "=" was an error since then the double integral would be 0.

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