Cartesian Polar Integral Converstion

Jun 2010
10
0
PLZ HELP! appreciations

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

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Prove It

MHF Helper
Aug 2008
12,897
5,001
I don't see any reason to change this integral to a polar integral...
 

Jhevon

MHF Helper
Feb 2007
11,681
4,225
New York, USA
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.
 
Jun 2010
10
0
it is required bythe book ..........i wouldnt want to change it ether if i wasnt required to
 
Jun 2010
10
0
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
Feb 2007
11,681
4,225
New York, USA
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

\(\displaystyle 0 < r < 6 \csc \theta\)
 
Jun 2010
10
0
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!