Integrate Multivariable Function using polar co-ords

Hello,

Trying to solve this,

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

Over the reigon A, where A is

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

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!

Re: Integrate Multivariable Function using polar co-ords

We have:

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

So, the integral is $\displaystyle I=\int_0^{\pi/2}d\theta\int_0^{\sqrt{\pi/2}} \rho\;(\cos \rho^2)\;d\rho $

Re: Integrate Multivariable Function using polar co-ords

Thanks for the response, could you please explain why

Quote:

Originally Posted by

**FernandoRevilla** $\displaystyle 0\leq y\leq \sqrt{\pi/2-x^2}$

?

Thank you!

Re: Integrate Multivariable Function using polar co-ords

Sketch the quarter of circle: fixing $\displaystyle x\in [0,\sqrt{\pi/2}]$ , $\displaystyle y$ varies from $\displaystyle 0$ to $\displaystyle \sqrt{\pi/2-x^2}$ .

Re: Integrate Multivariable Function using polar co-ords

Quote:

Originally Posted by

**MattWT** Hello,

Trying to solve this,

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

Over the reigon A, where A is

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

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 $\displaystyle \int_a^b\int_c^d f(x,y)dxdy$ **only** if the region of integration is a rectangle!