Surface area of sphere with cylindrical cutout

Question

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?

Answer 1

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.

Answer 2

Your integral in polar form is easy (not a trigonometry integral) if you simplify the integrand. Here is what I get including the correction mentioned by Zander and the factor of 2 you mentioned:\begin{align*} \text{surface area of spherical part}& = 2\int_0^{2\pi}\int_8^{11}\frac {11r}{\sqrt{121 - r^2}}dr\,d\theta\\ & = 22\cdot 2\pi\left[-\sqrt{121 - r^2}\right]_8^{11}\\ & = 44\pi\sqrt{57}.\end{align*} Then add the surface area of the cylinder. Its height is $2\sqrt{11^2 - 8^2} = 2\sqrt{57}$.

Answer 3

It is a direct computation based on already evaluated result/ formula for spherical cap. Let R^2 = h^2 + r^2, ( R = 11, r = 8 ). Area of spherical cap = 2 pi R h where R is sphere radius, h is total cap height, either for a full or for a truncated spherical segment.

4 pi R^2 - 2 pi R (2 R - 2 h) = 4 pi R h.

Interestingly also, remaining volume depends only upon h.