Use, using the result that for a simple closed curve C in the plane the area enclosed is:
A = (1/2)∫(x dy - y dx) to find the area inside the curve x^(2/3) + y^(2/3) = 4
∫P dx + Q dy = ∫∫ dQ/dx - dP/dy
The Attempt at a Solution
I solved the equation of the curve for x:
x = (4 - y^(2/3))^(3/2)
Also, from the original curve equation x^(2/3) + y^(2/3) = 4, when x = 0, y = +/- 8 because 4^(3/2) = 8.
But when I plug x = +/- (4 - y^(2/3))^(3/2) in for the x bounds and y = +/- 8 in for the y bounds in the resulting double integral
(1/2)∫∫ 2 dxdy
I have trouble integrating x = (4 - y^(2/3))^(3/2) it with respect to y.
Does anybody happen to know if there is a more correct way to solve this problem?
Thank you for your help!