I need to set this up as a double integral.

$\displaystyle x^2+y^2+z^2=1$

So the region of integration on the xy-plane should be $\displaystyle x^2+y^2\leq 1$ or,

$\displaystyle -1\leq x\leq 1$

$\displaystyle -\sqrt{1-x^2}\leq y\leq \sqrt{1-x^2}$

If I take the region in the first quadrant, I end up writing

$\displaystyle V=\int_R\int f(x,y)dA =4\int_{0}^1\int_0^{\sqrt{1-x^2}}\sqrt{1-x^2-y^2}dydx$

The answer I have in the book gives the constant as 8, instead of 4. I'm not understanding why this is. I'm integrating over 1/4 of the region defined by $\displaystyle x^2+y^2\leq 1$, so I'm just multiplying it by 4 to account for the rest of the region....