Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Hemisphere cut by cylinder

  1. Sep 17, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the area of the portion of the cylinder x^2 + y^2 = 2x that lies inside the hemisphere x^2 + y^2 + z^2 = 4, z [tex]\geq[/tex] 0. Hint: Project onto the xz-plane.

    2. Relevant equations

    I want to use the formula for surface area

    [tex]\int\int\frac{|\nabla f|}{|\nabla f\bullet\vec{p}|}dA[/tex]

    3. The attempt at a solution

    I'm going to consider only the surface in the first octant (for reasons of symmetry). I get

    [tex]\frac{|\nabla f|}{|\nabla f\bullet\vec{p}|} = \frac{1}{y}[/tex]

    hence:

    [tex]\int\int\frac{|\nabla f|}{|\nabla f\bullet\vec{p}|}dA = \int\int\frac{1}{\sqrt{2x-x^2}}dzdx[/tex]


    and using

    sqrt(4-x^2) and 0 as limits of integration for z

    and 2 and 0 as limits of integration for x, I get

    [tex]\int\sqrt{\frac{2+x}{x}}dx[/tex]

    (with 2 and 0 as limits of integration for x)

    The problem is that this integral doesn't evaluate to 4, which I know is the correct answer (I do get this result by evaluating the integral

    [tex]\int h ds[/tex]

    where h is the altitude of the cylinder and ds is the element of arc length on the circle x^2 + y^2 = 2x in the xy-plane)

    Could you please tell me where I'm going wrong?

    Many thanks in advance!
     
  2. jcsd
  3. Sep 18, 2010 #2
    I've just realized that there is a mistake in my limits of integration for z.

    Apart from 0, the other limit is sqrt(4 - (x^2 + y^2)), not sqrt(4 - x^2), and, since this z also belongs to the cylinder, z = sqrt(4 - 2x) and this leads to the correct result, i.e. 4.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook