
Volume of a Sphere
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 xyplane should be $\displaystyle x^2+y^2\leq 1$ or,
$\displaystyle 1\leq x\leq 1$
$\displaystyle \sqrt{1x^2}\leq y\leq \sqrt{1x^2}$
If I take the region in the first quadrant, I end up writing
$\displaystyle V=\int_R\int f(x,y)dA =4\int_{0}^1\int_0^{\sqrt{1x^2}}\sqrt{1x^2y^2}dydx$
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....


I think I've figured out what's wrong. The definition of volume for two variables requires that $\displaystyle f(x,y)> 0$ so my integral only accounts for the top half above the xyplane.