Double integrals using polar co-ordinates

1. Oct 28, 2012

Mdhiggenz

1. The problem statement, all variables and given/known data
∫∫e-(x^2+y^2) dA
R

Where R is the region enclosed by the circle x2+y2=1

First thing I did was graph the region where the function was enclosed. I saw that they didnt give a limitation to where the circle lied. So I automatically knew that d(theta) would lie on the interval 0≤theta≤2π.

Here is the part where things get a bit tricky for me, I need to find what dr is

So my thought process automatically told me to change x2+y2=1

Into polar coordinates.

Using the following relations x=rcos(theta); y=rsin(theta) and plugging in I get the relation

r^2=1

So I would think dr would lie on the interval from -1≤r≤1

However what the book says is that r lies on 0≤r≤1.

This I do not understand.

Thanks

Higgenz

2. Relevant equations

3. The attempt at a solution

2. Oct 28, 2012

Dick

r generally chosen to be a nonnegative number. Then 0<=r<=1 and 0<=theta<2*pi covers the unit disk once. If you let -1<=r<=1 then you have covered the unit disk twice. Changing theta->theta+pi and r->(-r) wouldn't change x and y.

3. Oct 28, 2012

Mdhiggenz

Wow thanks dick. However, what if for example we had a the same circle

X^2+y^2=4

Would then go from 0≤r≤2 ?

4. Oct 28, 2012

Dick

Sure. Same thing.

5. Oct 28, 2012

Mdhiggenz

Awesome thank you