# Domain for Polar Coordinate Part 2

## Homework Statement

f(x,y) = $e^{x^2+y^2}$
$x^{2}$ + $y^{2}$ ≤ R

## 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...

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Dick
Homework Helper
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.

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

HallsofIvy
Homework Helper
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
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