1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Another Surface Integral

  1. Oct 15, 2006 #1
    Okay - I thought that I figured this stuff out, but I didn't.
    The Problem
    When [tex]G(x, y, z) = (1-x^2-y^2)^{3/2}[/tex], and [tex]z = \sqrt{1-x^2-y^2}[/tex], evaluate the surface integral.
    My Work
    I keep trying this but I end up with the following integral that I cannot evaluate:

    [tex]
    \int_{-1}^{1} \!\!\! \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} (1-x^2-y^2)^{3/2}\sqrt{1+4x^2+4y^2} \,dx \,dy
    [/tex].

    Conversion to polar coordinates doesn't help much, either. How can I find this surface integral? (Ans: [itex]\frac{\pi}{2}[/itex])

    Thanks!
     
  2. jcsd
  3. Oct 15, 2006 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    How did you get that differential? [itex]z= \sqrt{1- x^2- y^2}[/itex] is the upper half of the sphere [itex]x^2+ y^2+ z^2= 1[/itex].

    Here's one way to treat it: think of the sphere as a 'level surface' of the function [itex]f(x,y,z)= x^2+ y^2+ z^2[/itex]. Then [itex]\nabla f= 2xi+ 2yj+ 2zk[/itex]. "Normalize that to the xy-plane by dividing through by the coefficient of k: [itex]\frac{x}{z}i+ \frac{y}{z}j+ 1[/itex] and take the length to get
    [tex]dS= \sqrt{\frac{x^2}{z^2}+ \frac{y^2}{z^2}+ 1}dxdy[/tex]
    [tex]= \frac{\sqrt{x^2+ y^2+ z^2}}{z}dxdy= \frac{1}{z}dxdy[/tex]
    Since [itex]z= \sqrt{1- x^2- y^2}[/itex], your integral becomes
    [tex]\int_{x= -1}^1\int_{y= -\sqrt{1-x^2}}^{\sqrt{1-x^2}}\frac{(1-x^2-y^2)^{3/2}}{(1-x^2-y^2)^\frac{1}{2}}dydx= \int_{x= -1}^1\int_{y= -\sqrt{1-x^2}}^{\sqrt{1-x^2}}(1- x^2- y^2) dydx[/tex]

    Another way: using spherical coordinates, with [itex]\rho= 1[/itex] gives parametric equations for the hemispher:
    [itex]x= cos(\theta)sin(\phi)[/itex], [itex]y= sin(\theta)sin(\phi)[/itex], and [itex]z= cos(\phi)[/itex], with [itex]0\le \theta \le 2\pi[/itex], [itex]0 \le \phi \le \frac{\pi}{2}[/itex].
    You can write the "position vector" of a point on the hemisphere as
    [tex] \vec{r}= cos(\theta)sin(\phi)\vec{i}+ sin(\theta)sin(\phi)\vec{j}+ cos(\phi)\vec{k}[/itex]
    Then the partial derivatives are
    [tex]\vec{r}_\theta = -sin(\theta)sin(\phi)\vec{i}+ cos(\theta)sin(\phi)\vec{j}[/tex]
    [tex]\vec{r}_\phi = cos(\theta)cos(\phi)\vec{i}+ sin(\theta)cos(\phi)\vec{j}- sin(\phi)\vec{k}[/tex]
    The "fundamental vector product" is the cross product of those:
    [tex]-cos(\theta)sin^2(\phi)\vec{i}- sin(\theta)sin^2(\phi)\vec{j}- sin(\phi)cos(\phi)\vec{k}[/tex]
    and its length gives the differential of surface area in those parameters:
    [tex]dS= sin^2(\phi)d\theta d\phi[/tex]

    Of course, [itex]1- x^2- y^2= 1- cos^2(\theta)sin^2(\phi)- sin^2(\theta)sin^2(\phi)= 1- sin^2(\phi)= cos^2(\phi)[/itex] so [itex](1-x^2-y^2)^\frac{3}{2}= cos^3(\phi)[/itex]

    In terms of those parameters your integral is
    [tex]\int_{\theta= 0}^{2\pi}\int_{\phi= 0}^{\frac{\pi}{2}}cos^3(\phi)sin^2(\phi) d\phi d\theta[/tex]
     
  4. Oct 15, 2006 #3
    I do not understand the "level surface," normalization, and "dividing through by the coefficient of k." Although mathematically I understand what you are doing, the concepts behind it confuse me.
    My textbook has not introduced gradients. It introduces divergence first, so technically I am not supposed to understand your gradient and normalization technique yet.

    I use the following integral to solve these types of problems:
    [tex]
    \int \!\!\! \int_R G(x, y, z) \sqrt{1 + \left( \frac{\partial f}{\partial x} \right ) ^2 + \frac{\partial f}{\partial y} \right ) ^2} \,dA
    [/tex]
    where [itex]R[/itex] is the projected surface (usually on the xy-plane), [itex]f[/itex] is one of the coordinates as a function of the other two (usually [tex]z = f(x, y)[/tex]), and [itex]dA[/itex] is the differential area element on the project (usually [tex]\,dx \,dy[/tex]). This integral is based on the idea that
    [tex]
    \hat{\textbf{n}} = \frac{-i\frac{\partial f}{\partial x} - j\frac{\partial f}{\partial y} + k}{\sqrt{1 + \left( \frac{\partial f}{\partial x} \right ) ^2 + \left( \frac{\partial f}{\partial y} \right )^2}}
    [/tex]

    EDIT: Never mind. Though I still don't understand your procedure, I really messed up in taking the partial derivatives of z. Therein was the problem.
     
    Last edited: Oct 15, 2006
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Another Surface Integral
  1. Another integration (Replies: 22)

  2. Another integral (Replies: 7)

  3. Another integral (Replies: 8)

  4. Another Integral (Replies: 15)

  5. Another integral (Replies: 3)

Loading...