# Integrate Multivariable Function using polar co-ords

• December 23rd 2011, 03:26 AM
MattWT
Integrate Multivariable Function using polar co-ords
Hello,

Trying to solve this,

$\int_a^b{\int_c^d{\cos(x^2+y^2)\,dx}\, dy}$

Over the reigon A, where A is

$x>=0, y>=0, x^2+y^2 <= \frac{pi}{2}$

I've attempted to work out initially what the limits a,b,c,d are. Would I be correct in thinking these were 0, sqrt(pi/2), 0, sqrt(pi/2)?

Then I'm struggling to convert these limits into polar form.

Thank you!
• December 23rd 2011, 03:36 AM
FernandoRevilla
Re: Integrate Multivariable Function using polar co-ords
We have:

$A \equiv \begin{Bmatrix} 0\leq x\leq \sqrt{\pi/2}\\0\leq y\leq \sqrt{\pi/2-x^2}\end{matrix} \equiv \begin{Bmatrix} 0 \leq \theta\leq \pi/2\\0\leq \rho\leq \sqrt{\pi/2}\end{matrix}$

So, the integral is $I=\int_0^{\pi/2}d\theta\int_0^{\sqrt{\pi/2}} \rho\;(\cos \rho^2)\;d\rho$
• December 23rd 2011, 03:40 AM
MattWT
Re: Integrate Multivariable Function using polar co-ords
Thanks for the response, could you please explain why

Quote:

Originally Posted by FernandoRevilla
$0\leq y\leq \sqrt{\pi/2-x^2}$

?

Thank you!
• December 23rd 2011, 03:45 AM
FernandoRevilla
Re: Integrate Multivariable Function using polar co-ords
Sketch the quarter of circle: fixing $x\in [0,\sqrt{\pi/2}]$ , $y$ varies from $0$ to $\sqrt{\pi/2-x^2}$ .
• December 23rd 2011, 05:00 AM
HallsofIvy
Re: Integrate Multivariable Function using polar co-ords
Quote:

Originally Posted by MattWT
Hello,

Trying to solve this,

$\int_a^b{\int_c^d{\cos(x^2+y^2)\,dx}\, dy}$

Over the reigon A, where A is

$x>=0, y>=0, x^2+y^2 <= \frac{pi}{2}$

I've attempted to work out initially what the limits a,b,c,d are. Would I be correct in thinking these were 0, sqrt(pi/2), 0, sqrt(pi/2)?

Then I'm struggling to convert these limits into polar form.

Thank you!

You have an integral of the form $\int_a^b\int_c^d f(x,y)dxdy$ only if the region of integration is a rectangle!