# Thread: Polar Double Integral Problem

1. ## Polar Double Integral Problem

Hi, I have been given the task of converting this integral:
$\int_0^1\int_{y^2}^{\sqrt{y}}(x+y) dx dy$

Into polar coordinates. I don't really know where to begin. I found that
$\sqrt{y}=x$ is equal to $tan \theta sec \theta$ and
$y^2=x$ is equal to $cot \theta csc \theta$,
But I don't know how that might help me. Any help?

Thanks,
Peter

2. ## Re: Polar Double Integral Problem

Originally Posted by flybynight
Hi, I have been given the task of converting this integral:
$\int_0^1\int_{y^2}^{\sqrt{y}}(x+y) dx dy$

Into polar coordinates. I don't really know where to begin. I found that
$\sqrt{y}=x$ is equal to $tan \theta sec \theta$ and
$y^2=x$ is equal to $cot \theta csc \theta$,
But I don't know how that might help me. Any help?

Thanks,
Peter
The first thing you may want to note is that the integrad

$f(x,y)=x+y$ is symmetric with respect to the line $y=x$

The line $y=x$ also symmetrically divides your region of integration. You can use this symmetry to your advantage. You only need to integrate over half of the region upto the line $y=x$ or from the line to the top.

This will be easier to convert

$2\int_{0}^{1} \int_{y^2}^{x}(x+y)dxdy$

to polar coordinates!

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