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Homework Help: 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:

    \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

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

  2. jcsd
  3. Oct 15, 2006 #2


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    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:
    \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
    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
    \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}}

    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
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