Solution for QID #44858: No matter which way you use integrals, the solution will alw | StudyHelpMe

No matter which way you use integrals, the solution will always come out to be ##pi*a*b##, where a and b are the semi-major axis and semi-minor axis. However, if you insist on using integrals, a good way to start is to split the ellipse into four quarters, find the area of one quarter, and multiply by four. To start with, we recognise that the formula for one quarter of an ellipse is ##y = b*sqrt((1-x^2)/a^2)## This quarter-ellipse is "centred" at ##(0,0)##. Its area is ##A = int_0^a(b*sqrt((1-x^2)/a^2))dx## So, naturally, the total area of the ellipse is ##A = 4int_0^a(b*sqrt((1-x^2)/a^2))dx##.