# Double integrals: cartesian to polar coordinates

1. Nov 27, 2015

### mmont012

1. The problem statement, all variables and given/known data
Change the Cartesian integral into an equivalent polar integral and then evaluate.

2. Relevant equations
x=rcosθ
y=rsinθ

I have:
∫∫r2cosθ dr dθ

The bounds for theta would be from π/4 to π/2, but what would the bounds for r be?

I only need help figuring out the bounds, not with the evaluating.

The answer for the problem is 36 (or so says the back of the textbook).

2. Nov 27, 2015

### blue_leaf77

The area of integration forms a right triangle subtending angle from 45 to 90 degree, so the limit for r would be a function of $\theta$. For a hint, as you sweep the triangle in between those two limiting angles, the projection $r \sin \theta$ is constant.

3. Nov 27, 2015

### HallsofIvy

For every theta, a line from the origin, making angle theta with the y-axis, to the line y= 6 is the hypotenuse of a right triangle with one leg of length 6. $cos(\theta)= \frac{6}{h}$.

4. Nov 30, 2015

### mmont012

Thank you both!