# Thread: Polar Double Integral Problem

1. ## Polar Double Integral Problem

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

Into polar coordinates. I don't really know where to begin. I found that
$\displaystyle \sqrt{y}=x$ is equal to$\displaystyle tan \theta sec \theta$ and
$\displaystyle y^2=x$ is equal to $\displaystyle cot \theta csc \theta$,
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:
$\displaystyle \int_0^1\int_{y^2}^{\sqrt{y}}(x+y) dx dy$

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

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

The line $\displaystyle y=x$ 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 $\displaystyle y=x$ or from the line to the top.

This will be easier to convert

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

to polar coordinates!

[img]22704[/img]