Find the volume of a sphere $x^2+y^2+z^2\leq 1$ contained between planes $z=1/2$ and $z=1/\sqrt2$ using cylindrical coordinates.

So the limits of $\theta$ would be $0$ to $2\pi$. Limits of $z$ would be the given planes. But why cant the limits of $r$ simply be $0$ to $1$? Is it because, if this were the case, we would be finding the volume of a cylinder and not a spherical type of object?

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    $\begingroup$ You are correct. The parameter $r$ can only go from $0$ to $\sqrt{1-z^2}$. $\endgroup$ – Brian Tung Apr 9 '15 at 1:38

Notice that $x^2+y^2=r^2$, thus $r^2+z^2=1$. Isolating $r$ gives $$r = \sqrt{1-z^2}.$$ If $0 \leq r \leq 1$ then you would obtain the cylinder that encloses sphere.


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