# Thread: Double Integration with polar co-ordinates

1. ## Double Integration with polar co-ordinates

Let D be the disk of radius 6 centered at the origin. Find the
double integral of the function f (x, y) = e^(x^2 +y^2) over D

When I do this i get a function that says e^r^2, and then I'm uncertain as to how to integrate this, I don't think I'm using polar co-ordinates correctly. Any suggestions/ideas?

2. recall dA = rdrdtheta

so you have re^r^2) which can be integrated with a simple substituion

3. what's the substitution? sorry my integration is a bit rusty...

4. you can let u = e^(r^2) or simply u = r^2

Let D be the disk of radius 6 centered at the origin. Find the
double integral of the function f (x, y) = e^(x^2 +y^2) over D

When I do this i get a function that says e^r^2, and then I'm uncertain as to how to integrate this, I don't think I'm using polar co-ordinates correctly. Any suggestions/ideas?
Calculus26 told you $\displaystyle dA=rdrd\theta$, probably you forgot the additional r.
Converting to polar:
$\displaystyle \int_0^{2\pi} \int_0^6 e^{r^2} r dr d\theta$.
Use the substitution $\displaystyle u=r^2$ to solve the inner integral.

6. Originally Posted by Ted
Calculus26 told you $\displaystyle dA=rdrd\theta$, probably you forgot the additional r.
Converting to polar:
$\displaystyle \int_0^{2\pi} \int_0^6 e^{r^2} r dr d\theta$.
Use the substitution $\displaystyle u=r^2$ to solve the inner integral.
Thanks for your reply, I think I was quite unclear in my question though, my problem lay with integrating a function that was e^u,(integrating the substitution), and when I tried to carry out the second integral, it did not have any variables with theta hence, I was unsure how to proceed...

7. Recall that: $\displaystyle \int a dx = ax + C$

i.e. after doing the inner integral, treat the result as a constant in terms of $\displaystyle \theta$.

8. Originally Posted by o_O
Recall that: $\displaystyle \int a dx = ax + C$

i.e. after doing the inner integral, treat the result as a constant in terms of $\displaystyle \theta$.
oh yes sorry, seems like a dumb question now, thanks!