Double integrals: cartesian to polar coordinates

[mmont012]

Homework Statement

Change the Cartesian integral into an equivalent polar integral and then evaluate.

Homework 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).

 

[blue_leaf77]

but what would the bounds for r be?

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.

 

[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. $cos(\theta)= \frac{6}{h}$.

 

[mmont012]

Thank you both!