Can someone double check my answer on a double integral?

Question

The outside integral has limits from y=0 to y=1. The inside integral goes from x=0 to x=squareroot(1-y^2). Inside the integral we have 9dxdy, so the equation is integrated with respect to x first, then y. After the inner integration, I have 9 * integral (0 to 1) of squareroot(1-y^2) dy. I found a formula in the back of my textbook that works for integrals of the form sauareroot(a^2 - u^2) du, where a is a constant (a=1 in my case). And my answer is 9 * (y/2 * squareroot(1-y^2) + 1/2 * sin-inverse(y)) evaluated from 0 to 1. It comes out to 9pi/4, but I think it's supposed to equal zero in the problem, so what did I do wrong? Thanks much!

Answer 1

That particular integral IS 9π/4. The problem is, as I explained in my other post, that your limits of integration for this particular problem should have been [-1, 1] and [-√(1-y²), √(1-y²)], not [0, 1] and [0, √(1-y²)], which naturally comes out to 9π. This is not zero - as I explained in the other post, your formula for the line integral was wrong.

Answer 2

The answer is 9pi/4, you are correct!