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.