Need help with surface area of this sphere

**ihaveamathproblem** · March 26th 2008, 02:29 PM

i need to determine the surface area of the portion of the sphere $x^2 + y^2 + z^2 = 16$ that lies between the planes $z=1$ and $z=2$
i have taken the derivatives with respect to x and y and have switched to polar coordinates and now i have the integral:
$\int \int {{(4/\sqrt(16-r^2)dA}}$
I am not sure what the limits of the integrals will be. But i am pretty sure the leftmost integral has the limits 0 to $2\pi$ . can someone help me figure out the limits of the integrals? thanks.

**galactus** · March 26th 2008, 02:47 PM

If you sub in z=1 and z=2 into the sphere equation you get:

$r^{2}=15, \;\ r^{2}=12$

So, you get:

$\int_{0}^{2\pi}\int_{\sqrt{12}}^{\sqrt{15}}\frac{4 r}{\sqrt{16-r^{2}}}drd{\theta}$

$=4\int_{0}^{2\pi}d{\theta}=8{\pi}$

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