Evaluate a simple double integral

[theneedtoknow]

Evaluate a "simple" double integral

Homework Statement

Evaluate the double integral of f(x,y) = square root (1 - x^2 - y^2) over the disk centred at the origin of radius 1

Homework Equations

The Attempt at a Solution

So the disc of radius one has boundaries x^2 + y^2 = 1
i am integrating it in the order dydx
the disk is bound above by y = root (1 - x^2) and below by y = -root (1 - x^2)

So the inner part of the double integral would be
Integral from -root (1-x^2) to root (1-x^2) of root (1 - x^2 - y^2) dy

in a list of integrals i found that if you have an integral in the form root (a^2 - u^2), the antiderivative is u/2 * root (a^2 - u^2) + a^2/2 * sin^-1 (u/a)

so if i set a^2 = 1-x^2 and u^2 = y^2, my integral has exactly this form, so the solution should be
root(1-x^2)/ 2 * root (1 - x^2 - (1 - x^2)) + (1-x^2)/2 * sin^-1 (root (1-x^2)/ root (1-x^2) - [ (-root(1-x^2) / x * (root (1 - x^2 - (1 - x^2)) + (1-x^2)/2 * sin ^-1 (-root(1-x^2) / root (1-x^2)

Which simplifies pretty cleanly to

(1-x^2) / 2 * (sin^-1 (1) - sin^-1 (-1)

= (1-x^2)/2 * (-Pi)
= -Pi/2 + Pix^2/2

Now for the outer integral, i have to integrate this from x = -1 to x = 1

and I get
Integral of -Pi/2 + Pix^2/2 = -Pi x/2 + Pi x^3 /6
evaluating this from -1 to 1 i get
- Pi / 2 + Pi/6 - (Pi / 2 - Pi / 6) = -2Pi / 3

But this gives me a negative value and I thought that the integral should be positive since it is essentally calculating the volume below the curve z = root (1 - x^2 - y^2)...and that function is positive for the entire disk of radius 1 , so i don't see why i am getting a negative number...what in the world is going on :)

 

[VeeEight]

Using polar coordinates from the start will simplify the question

 

[theneedtoknow]

Yeah but we have not yet learned polar coordinates :) so I have no idea how to use them