Integration - converting to polar coordinates

[willmoore21]

Hi guys

Have a double integration problem with converting to polar. I know the basics such as x=rcos(theta) etc, but am unsure of how to deal with things that are a little more tricky.

One of the limits is sqrt(1-x^2). The integral itself is x^2+y^2 dydx between 0,1, and 0,sqrt(1-x^2)

I know that x=rcos(theta), so I thought this would be equal to sqrt(1-r^2cos^2(theta)), but in the answers posted at the university, the limit is 2cos(theta), so how do I get there?


Apologies for no latex, I can't get square roots to work, just comes out with loads of <i>'s and stuff.

Thanks

Will

 

[galactus]

Is this your given integral?:

$\displaystyle \displaystyle \int_{0}^{1}\int_{0}^{\sqrt{1-x^{2}}}(x^{2}+y^{2})dydx$

If so, using polar we have:

$\displaystyle \displaystyle \int_{0}^{\frac{\pi}{2}}\int_{0}^{1}r^{3}drd\theta$

I see no reason for a $\displaystyle 2\cos\theta$. Are you sure you looked at the correct problem?.

Something like $\displaystyle (x-1)^{2}+y^{2}=1$ may result in $\displaystyle r=2cos\theta$. But this would be another problem.

Latex code varies from site to site. A radical will work if you use \sqrt{1-x^{2}} surrounded by the tags

 

[willmoore21]

My lecturer is on vacation for a week so will check to see if this is a mistake, he is very prone to this sort of thing.

Thanks for answer galatucus, but it's how you got there I'm interested in. I don't see how it works, can you point me in the right direction?

Will

 

[galactus]

$\displaystyle x^{2}+y^{2}=r^{2}$

So, with this $\displaystyle r^{2}$ and the r that tags along when we use polar, we get $\displaystyle r^{3}$.

The upper limit is the upper half of the unit circle, $\displaystyle \sqrt{1-x^{2}}$. So, this makes r = 0 to 1 and

theta = 0 to Pi/2.