Convert triangle vertices to double integral polar coordiantes

ramses07
Messages
11
Reaction score
0

Homework Statement



integrate

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

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

Homework Equations



x= rcosO, y = rsinO

x^2+y^2=r^2

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 can't seem to find the answer.
 
Physics news on Phys.org
I think your bounds for theta should be pi/4 to pi/2
 
how do you know what the bounds are?
 
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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top