Cartesian Polar Integral Converstion

**de20** · June 27th 2010, 07:07 PM

PLZ HELP! appreciations

change the cartesian integral into polar integral then evaluate the polar integral
$Cartesian Polar Integral Converstion-math.jpg$ 0<y<6 0<x<y xdxdy

**Prove It** · June 27th 2010, 07:28 PM

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

**Jhevon** · June 27th 2010, 07:35 PM

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

$x \mapsto r \cos \theta$

$y \mapsto r \sin \theta$

and

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

Hopefully you can use this to figure it out.

**de20** · June 27th 2010, 07:37 PM

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

**de20** · June 27th 2010, 08:11 PM

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

**Jhevon** · June 27th 2010, 08:24 PM

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

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

Switch to polar:

$0 < r \sin \theta < 6$

which means

$0 < r < 6 \csc \theta$

**de20** · June 27th 2010, 08:49 PM

Originally Posted by jhevon

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

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

switch to polar:

$0 < r \sin \theta < 6$

which means

$0 < r < 6 \csc \theta$

thank you!

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