Domain for Polar Coordinate Part 2

[DrunkApple]

Homework Statement

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

Homework Equations

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...

 

[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.

 

[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

 

[HallsofIvy]

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

Use \int for a \int in LaTeX.