# Surface area of sphere with cylindrical cutout

Compute the total surface area of the remaining part of a solid ball of radius 11 after a cylindrical hole of radius 8 is drilled through the center of the ball.

Change to cylindrical coordinates:

$x^2+y^2+z^2=11^2$ , thus $z=\sqrt{121-x^2-y^2}=\sqrt{121-r^2}$

$SA=\int\int_R \sqrt{1+(\partial{z}/\partial{x})^2+(\partial{z}/\partial{y})^2}dA$

Here $(\partial{z}/\partial{x})=(121-x^2-y^2)^{-1/2}(-2x)=(-2x)(121-r^2)^{-1/2}$

and $(\partial{z}/\partial{y})=(121-x^2-y^2)^{-1/2}(-2y)=(-2y)(121-r^2)^{-1/2}$

$(\partial{z}/\partial{x})^2+(\partial{z}/\partial{y})^2=(4r^2)/(121-r^2)$

$SA=\int_0^{2\pi}\int_8^{11}\sqrt{1+(\partial{z}/\partial{x})^2+(\partial{z}/\partial{y})^2}rdrd\theta$

$=\int_0^{2\pi}\int_8^{11}\sqrt{1+(4r^2)/(121-r^2)}rdrd\theta$

which is some trig integral that I would rather not do, so at this point I passed it off to Wolfram Alpha's double integral widget http://www.wolframalpha.com/widget/widgetPopup.jsp?p=v&id=f45d619ba0a669dc4191493d49ac4898&title=Double+Integral+Calculator&theme=blue using $x$ as $r$ and $y$ as $\theta$, and using the following equation for $f(x,y)$:

x(1+4x^2/(121-x^2))^(1/2)

, using the "dxdy" method, $x$ going from 8 to 11, $y$ going from 0 to 2*pi. This gives me the answer 978.405, which of course is just the top ($z$-positive) half of the surface, so multiply by 2 to get 1956.81.

But, homework says answer isn't right. It all looks right to me, where did I go wrong?

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No need for integration. Use these formulas en.wikipedia.org/wiki/Spherical_cap –  tom Mar 31 at 10:44
Where you've gone wrong in your work so far is that $\partial z/\partial x$ and $\partial z/\partial y$ are missing factors of $1/2$.