# Thread: Integrating using Polar Coordinate

1. ## Integrating using Polar Coordinate

Using polar coordinates, evaluate the double integral over r sin(x^2+y^2)dA where R is the region 4<x^2+y^2<36

Need to know where to start, or an example? I keep getting 8.something and it's incorrect.

2. ## Re: Integrating using Polar Coordinate

Originally Posted by Bracketology
Using polar coordinates, evaluate the double integral over r sin(x^2+y^2)dA where R is the region 4<x^2+y^2<36
The region you are integrating over is the region enclosed by (and not including) the circles centred at the origin of radii 2 and 6. So clearly your r bounds are 2 and 6, and since you are sweeping over the whole circle, your theta bounds are 0 and 2pi.

3. ## Re: Integrating using Polar Coordinate

I get that. But when I integrate and plug in, I get the wrong answer. So I integrate sin(r) from r=2 to 6? that give me -cos(r) from r=2 to 6. Then, integrate -cos(2)+cos(6) from theta=0 to 2pi?

This gives me 8.647 which is incorrect.

4. ## Re: Integrating using Polar Coordinate

I think that this is the expression that you should integrate:

$r\sin\left(x^2+y^2\right)\mathrm{d}A=r\sin\left(r^ 2)r\mathrm{d}r\mathrm{d}\theta$

5. ## Re: Integrating using Polar Coordinate

Originally Posted by corsica
I think that this is the expression that you should integrate:

$r\sin\left(x^2+y^2\right)\mathrm{d}A=r\sin\left(r^ 2)r\mathrm{d}r\mathrm{d}\theta$
Thanks for the imput, but how do I integrate rsin(r^2)*r? Or is the integral suppose to be rsin(r^2)?

I think you meant rsin(r^2), because that would make sense.

6. ## Re: Integrating using Polar Coordinate

The width of an infinitessimal element expressed in polar parameters is $r\mathrm{d}\theta$, and its length is $\mathrm{d}r$. So the area of the infinitessimal element is:
$\mathrm{d}A=r\mathrm{d}r\mathrm{d}\theta$
That's where the 'extra' factor $r$ comes from.

As far integrating it, I don't have time to check it myself right now, but integration by parts might be the way to go.

7. ## Re: Integrating using Polar Coordinate

I have a feeling your original integrand must have been \displaystyle \begin{align*} \sin{\left(x^2 + y^2\right)} \end{align*}, not \displaystyle \begin{align*} r\sin{\left(x^2 + y^2\right)} \end{align*} seeing as it doesn't make sense to have mixed polars with cartesians in the first place. Also, \displaystyle \begin{align*} \int{r^2\sin{\left(r^2\right)}\,dr} \end{align*} doesn't have a closed form answer, whereas the first will.