# Cartesian Polar Integral Converstion

#### Prove It

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

#### Jhevon

MHF Helper
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.

#### de20

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

#### de20

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

MHF Helper
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

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

#### de20

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!