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Homework Help: Computing a surface integral with polar coordinates

  1. May 16, 2012 #1
    1. The problem statement, all variables and given/known data
    Show that ##\iint_{S}(x^2 + y^2)d\sigma = \frac{9\pi}{4}##
    where ##S = \{(x,y,z): x > 0, y > 0, 3 > z > 0, z^2 = 3(x^2 + y^2)\}##

    2. Relevant equations
    ##\iint_{S}f(x,y,z)d\sigma = \iint_{R}f(r(x,y))\sqrt{[r_x(x,y)]^2 + [r_y(x,y)]^2 + 1}##
    where ##r : R → ℝ^3, R \in ℝ^2## parametrizes the surface S.

    3. The attempt at a solution

    ##\iint_{S}(x^2 + y^2)d\sigma = \iint_{R}(x^2 + y^2)\sqrt{\frac{9^2}{3(x^2 + y^2)} + \frac{9^2}{3(x^2 + y^2)}+ 1}dxdy##
    ## = \int_{0}^{2\pi}\int_{0}^{3}r^2\sqrt{\frac{9r^2}{3r^2} + 1} drd\theta## (polar coordinates)
    ##= \int_{0}^{2\pi}\int_{0}^{3}2r^2drd\theta = 36\pi##

    This is obviously wrong. I'm guessing I have the wrong domain restrictions
    ##R = \{ (r, \theta) : 0 < \theta < 2\pi, 0 < r < 3\}##
    If so, I can't figure out how to get the correct restriction.
  2. jcsd
  3. May 16, 2012 #2


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    You should draw the graph to find the limits. S is the surface of a cone in the first octant.
    [tex]z=\pm \sqrt{3x^2+3y^2}[/tex]
    But since we are considering the cone in the 1st octant, where [itex]z≥0[/itex], thus:
    Find [itex]z_x[/itex] and [itex]z_y[/itex] as you'll be projecting the surface S onto the x-y plane.

    The surface area in terms of Cartesian coordinates:
    [tex]\iint_{S}(x^2 + y^2)d\sigma = \iint_{Ω} (x^2 + y^2) \sqrt {(z_x)^2+(z_y)^2+1} \,.dxdy=\iint_{Ω} (x^2 + y^2) \sqrt {\frac{12x^2+12y^2}{3x^2+3y^2}} \,.dxdy=2\iint_{Ω} (x^2 + y^2) \,.dxdy[/tex] where Ω is the projection of S onto the x-y plane.

    To find the equation of the points of intersection of the plane z = 3 and the cone, substitute z = 3 into the equation of the cone.
    The projected area in terms of polar coordinates:
    [tex]2 \iint_Ω r^2 \,.rdrd\theta[/tex]
    Now, using the graph of the region Ω, find the limits. See attached graph.

    But from the projection onto the x-y plane, you only need a quarter of Ω (since the section of surface S required is found in the 1st octant), so [itex]0\le \theta \le \pi /2[/itex]

    Attached Files:

    Last edited: May 16, 2012
  4. May 16, 2012 #3
    Alright, I got it. Thanks for your help.
    I think I get the gist of the radians restriction; so if we have just x > 0, y > 0, but no z > 0 would it be ##R = \{(r, \theta); 0 < r < \sqrt{3}, 0 < \theta < \pi\}##?
  5. May 16, 2012 #4


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    You are correct except for the limits of [itex]\theta[/itex], which should be:
    [tex]0 < \theta < \pi /2[/tex] as the surface area that this problem requires is found in the first octant, or in case of no z-axis, then it's the 1st quadrant.
  6. May 16, 2012 #5
    Okay so if it's just {y > 0, but no x > 0, z > 0} OR {x > 0, but no y > 0, z > 0}, two quadrants of the space, then would it be ##0 < \theta < \pi##?
  7. May 16, 2012 #6


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    Draw the Cartesian axes in a 3D coordinate system; x, y and z, perpendicular to each other. Label them with simple values {-1,0,1} for each axis. For example, if x > 0, then it means you are only considering the positive x axis. As you are given more limits, you will be able to progressively pinpoint the exact octant required.

    Likewise, if y > 0, it means the graph is along the positive y axis. You can apply this simple means of knowing the required section of graph by determining which octant or quadrant is related to the problem.

    In your original problem, {x > 0, y > 0, 0 < z < 3}. Consider x > 0 and y > 0. This automatically places the required section of the graph in the first quadrant, but since your problem involves 3D coordinates, you now have an additional upper and lower dimension (the z axis). If z > 0, then the section is found in the 1st quadrant but above the z axis, meaning in the positive (upper) direction of the axis, in other words, the 1st octant. If z < 0, then it's obviously below the x-y plane, meaning, the required section of the graph is along the negative z axis.
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