Sketch the solid whose volume is given by the iterated integ

Sketch the solid whose volume is given by the iterated integral below, and find the volume.

Solution

I = integral of (1-x^2-y^2)dydx
integral of a^2-y^2 = y/2 * a^2-y^2 + a^2/2 *sin^-1(y/a)
here a^2 = 1-x^2
so by putting the formula & integrating wrt y, we get
I = integral of y/2 * (1-x^2-y^2) + (1-x^2)/2 *sin^-1(y/(1-x^2))dx
applying limit of y
I = integral of (1-x^2)/2 * (1-x^2-1+x^2) + (1-x^2)/2 *sin^-1((1-x^2)/(1-x^2))dx
I = integral of (1-x^2)/2 * /2dx
I = /4 integral of (1-x^2)dx
integrating wrt x
I = /4 (x-x^3/3)
applying limits of x
I = /4 (1-1/3)
I = /6