1. ## double integral question

Ok. I have a quick question about the limits of integration for a problem:
*I am going to use @ for theta.*

Integrate f(x,y) = x + y over the region R bounded by r = cos(@)

I converted the function to a polar function and got r cos(@) + r sin(@) and a r dr d@ making it:

r^2[cos(@) + sin(@)]dr d@

now for the limits of integration I am using 0 to cos(@) and 0 to 2 pi Is this correct or should I use 0 to 1 for the dr since cos can only go from 0 to 1? I am thinking I am correct with my first method but I just wanted to check. Also if I made any other mistakes feel free to point those out as well.

2. Originally Posted by billbarber
Ok. I have a quick question about the limits of integration for a problem:
*I am going to use @ for theta.*

Integrate f(x,y) = x + y over the region R bounded by r = cos(@)

I converted the function to a polar function and got r cos(@) + r sin(@) and a r dr d@ making it:

r^2[cos(@) + sin(@)]dr d@

now for the limits of integration I am using 0 to cos(@) and 0 to 2 pi Is this correct or should I use 0 to 1 for the dr since cos can only go from 0 to 1? I am thinking I am correct with my first method but I just wanted to check. Also if I made any other mistakes feel free to point those out as well.

$r=\cos \theta\,,\,\,0\leq\theta\leq \pi$ is the same as the circle $\left(x-\frac{1}{2}\right)^2+y^2=\frac{1}{4}$ ...

Tonio

3. so are you saying it would be easier to just convert the bound r = cos(@) to rectangular coordinates?

4. or are you saying that the theta goes from 0 to pi and the r goes from 0 to 1/2?

5. Originally Posted by billbarber
or are you saying that the theta goes from 0 to pi and the r goes from 0 to 1/2?

No: r goes as cos, as given: $r=\cos\theta$, and $\theta$ ranges between 0 and $\pi$

Tonio