Thread: Cartesian Polar Integral Converstion

PLZ HELP! appreciations

change the cartesian integral into polar integral then evaluate the polar integral
0<y<6 0<x<y xdxdy

I don't see any reason to change this integral to a polar integral...

I agree with Prove It. But if this is just your professor being annoying and wants you to use Polar coordinates even when it is unnecessary, you can recall that for Polar coordinates

$\displaystyle x \mapsto r \cos \theta$

$\displaystyle y \mapsto r \sin \theta$

and

$\displaystyle dx~dy \mapsto r ~dr~d \theta$

Hopefully you can use this to figure it out.

it is required bythe book ..........i wouldnt want to change it ether if i wasnt required to

can u atleast show me the bounds of integrations pleasae? i was wondering if i can have 0<r<root72 sinthata or 0<r<rsin thata

Hopefully you can see that $\displaystyle \frac {\pi}4 \le \theta \le \frac {\pi}2$

Now, you know that $\displaystyle 0 < y < 6$

Switch to polar:

$\displaystyle 0 < r \sin \theta < 6$

which means

$\displaystyle 0 < r < 6 \csc \theta$

Originally Posted by jhevon

hopefully you can see that $\displaystyle \frac {\pi}4 \le \theta \le \frac {\pi}2$

now, you know that $\displaystyle 0 < y < 6$

switch to polar:

$\displaystyle 0 < r \sin \theta < 6$

which means

$\displaystyle 0 < r < 6 \csc \theta$

thank you!