Convert triangle vertices to double integral polar coordiantes

1. Apr 19, 2010

ramses07

1. The problem statement, all variables and given/known data

integrate

f(x,y) = sqrt(x^2+y^2)

over triangle with vertices (0,0) (0,sqrt2) (sqrt 2, sqrt 2)

2. Relevant equations

x= rcosO, y = rsinO

x^2+y^2=r^2
3. The attempt at a solution

im supposed to use a double integral converted to polar coordinates,
so i used the bounds int. 0 to pi/4 int. 0 to sqrt 2 sec (r^2) drdO

are these the correct bounds? because i cant seem to find the answer.

2. Apr 19, 2010

cragar

I think your bounds for theta should be pi/4 to pi/2

3. Apr 19, 2010

ramses07

how do you know what the bounds are?

4. Apr 19, 2010

cragar

because the line y=x cuts it at a 45 degree angle and x=0 goes up to 90 so it goes from 45 to 90
like cutting a wedge out of a circle , but the radius is different , i think your bounds for the radius are correct.