Converting to polars

willmoore21

Junior Member
Joined
Jan 26, 2012
Messages
75
Hi guys,

I've always had problems converting to polars, mainly because I don't think I understand the fundamentals.

I really don't understand why things like sqrt(1-x^2) = 1 and things like that.

I have a revision question I would like help with, and maybe some links to any useful websites that you guys know.

Question

Use polar coordinates to evaluate

int.int(overR) sqrt(9-x^2-y^2) dA. where R is the region in the first quadrant of a circle within domain x^2+y^2=9.

Now I know that when I sketch this, r goes from 0 to pi/2, but I've no idea how to work out the second limit.

Apologies for no latex.

Will
 
Hi guys,

I've always had problems converting to polars, mainly because I don't think I understand the fundamentals.

I really don't understand why things like sqrt(1-x^2) = 1 and things like that.

I have a revision question I would like help with, and maybe some links to any useful websites that you guys know.

Question

Use polar coordinates to evaluate

int.int(overR) sqrt(9-x^2-y^2) dA. where R is the region in the first quadrant of a circle within domain x^2+y^2=9.

Now I know that when I sketch this, r goes from 0 to pi/2, but I've no idea how to work out the second limit.

Apologies for no latex.

Will
No, r does NOT go from "0 to pi/2", the angle \(\displaystyle \theta\) goes from 0 to \(\displaystyle \pi/2\). Since x^2+ y^2= 9 is a circle of radius 3, r goes from 0 to 3. And don't forget that, in polar coordinates, the "differential of area" is \(\displaystyle r drd\theta\).
 
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