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?
Best Answer
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 $1/2$.
Furthermore, what you're trying to calculate above is the area of the exterior of the sphere that remains after drilling out the hole (and the result you get is larger than total surface area, so cannot be correct). But to answer the question you'll also need to include the area of the surface of the interior of the sphere exposed by drilling out the hole.