# Evaluating a repeated integral by changing to polar co-ordinates

• Dec 10th 2008, 11:18 AM
chella182
Evaluating a repeated integral by changing to polar co-ordinates
Okay, so I missed my lectures on this and I haven't a clue (Crying) I have my friend's assignment here but she's gone from step to step and I totally don't understand how. The question is:

Evaluate by changing to polar co-ordinates
$\int_{0}^{1}\int_{0}^{\sqrt{1-x^2}}\sin(\frac{\pi(x^2+y^2)}{2})dydx$

My friend has then wrote down $x=r\cos\theta$ and $y=r\sin\theta$ then jumped to...

$\int_{0}^{\pi/2}\int_{0}^{1}\sin(\frac{\pi r^2}{2})r drd\theta$

...and I don't know how. Can anyone go through it explaining the steps to me?
• Dec 10th 2008, 11:47 AM
running-gag
Hi

Have a look to the scheme below
$\int_{0}^{1}\int_{0}^{\sqrt{1-x^2}}f(x,y)dydx$ means that you integrate f on the domain defined by $0 \leq x \leq 1$ and for each value of x $0 \leq y \leq \sqrt{1-x^2}$

When you plot the function $y = \sqrt{1-x^2}$ you can see that it is a quarter of a circle.
x between 0 and 1, and y between 0 and $\sqrt{1-x^2}$ defines the quarter of the disk as area of integration (see red line).

If you want to change to polar co-ordinates, you have to sweep the same domain using $\theta$ and r.

You can see that you can manage this by choosing $0 \leq \theta \leq \frac{\pi}{2}$ and for each value of $\theta$ $0 \leq r \leq 1$

Now you have to substitute $x=r\cos\theta$ and $y=r\sin\theta$
$\sin(\frac{\pi(x^2+y^2)}{2}) = \sin(\frac{\pi r^2}{2})$

and $dx dy$ by $r drd\theta$

to get finally

$\int_{0}^{1}\int_{0}^{\sqrt{1-x^2}}\sin(\frac{\pi(x^2+y^2)}{2})dydx = \int_{0}^{\pi/2}\int_{0}^{1}\sin(\frac{\pi r^2}{2})r drd\theta$

http://imageshack-france.com/out.php..._Integral3.JPG
• Dec 10th 2008, 01:14 PM
chella182
Thankyou :) it's a lot clearer than it was, anyway. I just didn't want to blindly copy the work without actually knowing what I was copying / if what I was copying was actually right. I don't get how she's managed to get between those two steps with no working out in between like haha!
• Jan 16th 2009, 02:43 AM
chella182
I get that to change to polar co-ords that $x=r\cos{\theta}$ and $y=r\sin{\theta}$.
But then, on my assignment I've written...

$\sqrt{1-x^2}$ makes a quarter circle, so $0 \leq \theta \leq \frac{\pi}{2}$ and $0 \leq r \leq 1$

...but I totally don't get where I've pulled that from (Crying)
• Jan 16th 2009, 04:44 AM
mr fantastic
Quote:

Originally Posted by chella182
I get that to change to polar co-ords that $x=r\cos{\theta}$ and $y=r\sin{\theta}$.
But then, on my assignment I've written...

$\sqrt{1-x^2}$ makes a quarter circle, so $0 \leq \theta \leq \frac{\pi}{2}$ and $0 \leq r \leq 1$

...but I totally don't get where I've pulled that from (Crying)

What part of the explanation you've been given in this thread do you not understand?

Note: $y = \sqrt{1 - x^2} \Rightarrow y^2 = 1 - x^2 \Rightarrow x^2 + y^2 = 1$ which is a circle. You only want the part of the circle given by 0 < x < 1 and y > 0, that is, the part of the circle lying in the first quadrant.
• Jan 17th 2009, 02:12 AM
chella182
I posted this ages ago. I didn't understand where the 0 and 1 came from on the outer limits. I kind of do now. Flunked this question on the exam still though (Crying)