Volume of a Sphere

**adkinsjr** · October 22nd 2010, 04:13 PM

I need to set this up as a double integral.

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

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

$-1\leq x\leq 1$

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

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

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

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 $x^2+y^2\leq 1$ , so I'm just multiplying it by 4 to account for the rest of the region....

**Also sprach Zarathustra** · October 22nd 2010, 04:24 PM

http://www.math.wvu.edu/~hjlai/Teach...df/Tip3-33.pdf

The last part...

**adkinsjr** · October 22nd 2010, 04:43 PM

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

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