1. ## Double Polar Integral

Evaluate the double integral: (Two integration symbols over area D) 4x - 5y dA

Where D is the region enclosed by the half-annulus bordered in red for 2 pi/3 <= Theta <= 5 pi/3. The inside radius is 4 and the outside radius is 8.

I set up the integral with the inside integral having an interval from 4 to 8 and the outside integral having an interval from 2 pi / 3 to 5 pi /3. I converted the x in the equation to r * cos(theta) and all the y into r * sin(theta), and multiplied the entire equation by r, making the final equation 4 * r^2 * cos(theta) - 5 * r^2 * sin(theta). Then, I integrated the double integral using a calculator and got -287.945 as my answer. This was not the correct answer.

2. What is your closed solution for

$I=\displaystyle\int_{2\pi/3}^{5\pi/3}(4\cos \theta-5\sin \theta)\;d\theta \displaystyle\int_{4}^{8}\rho^2\;d\rho$ ?

3. If you're talking about the inside first integral, I got 597.333 * (cos(theta) - 1.25 * sin(theta))

4. Computing by hand, I get:

$I=\dfrac{448(5-4\sqrt{3})}{3}$

5. That's the same answer I got. The computer still won't accept it though.