# Domain for Polar Coordinate Part 2

1. Nov 13, 2011

### DrunkApple

1. The problem statement, all variables and given/known data
f(x,y) = $e^{x^2+y^2}$
$x^{2}$ + $y^{2}$ ≤ R

2. Relevant equations

3. The attempt at a solution
I believe this is a circle.

$f^{2pi}_{0}$$^{sqrt(R)}_{-sqrt(R)}$$e^{x^2+y^2}$*r*dr*dθ

= $f^{2pi}_{0}$$f^{sqrt(R)}_{-sqrt(R)}$$e^{r^2}$*r*dr*dθ

after u substitution...

= $\frac{1}{2}$$f^{2pi}_{0}$$f^{R}_{R}$$e^{u}$*du

Does this makes sense?

Of course not... r domain are same...

2. Nov 13, 2011

### Dick

Polar coordinates are defined on the domain r>=0 and 0<=theta<2*pi. r is nonnegative. Your lower limit for r should be 0.

3. Nov 13, 2011

### DrunkApple

ohhh so r can never be negative so they are always equal to or bigger than 0?
ok I got it.
Thank u

So the correct one is

= $\frac{1}{2}$$f^{2pi}_{0}$$f^{R}_{0}$$e^{u}$*du

4. Nov 14, 2011

### HallsofIvy

Use \int for a $\int$ in LaTeX.