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Calc 3, area of an ellipsoid slice

  1. Oct 10, 2011 #1
    This isn't homework or anything, I just want to understand the question better.

    1. The problem statement, all variables and given/known data
    calcproblem.jpg



    3. The attempt at a solution

    I'm honestly not sure where to go with this. Is this an integral problem? As I understand it I'm finding the area of a slice, not a volume of the whole ellipsoid.

    so z = c which = 3...
    the problem then becomes:
    x[itex]^{2}[/itex] + [itex]\frac{y^{2}}{4}[/itex] = 0

    Does the last description mean that the answer is A= ([itex]\pi[/itex])(1)(2) ?
    It doesn't seem like it should be that simple. lol
     
  2. jcsd
  3. Oct 10, 2011 #2

    Mark44

    Staff: Mentor

    No, because x2 + y2/4 = 0 is not the equation of an ellipse.

    Notice that there aren't many points that satisfy this equation.

    Have you drawn a sketch of the ellipsoid? A sketch would help illuminate what's going on here.
     
  4. Oct 10, 2011 #3

    SammyS

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    Staff Emeritus
    Science Advisor
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    No, it's not an integral problem.

    The only real solution for [itex]\displaystyle x^2+\frac{y^2}{4}=0[/itex] is x = ? = y .

    Set z = c , then solve for [itex]\displaystyle x^2+\frac{y^2}{4}\,,[/itex] then divide both sides by the right hand side.
     
  5. Oct 10, 2011 #4

    Mark44

    Staff: Mentor

    Where does it say that the plane is z = 3?
     
  6. Oct 10, 2011 #5
    calcprobgraph.jpg




    But isn't 9 supposed to be c[itex]^{2}[/itex]?

    so really I have:
    x[itex]^{2}[/itex] + [itex]\frac{y^{2}}{4}[/itex] + [itex]\frac{c^{2}}{c^{2}}[/itex] = 1
    which just leads me back to the previous equation where the only x and y values would be zero.
    or am I just making a false assumption?

    am I supposed to be doing this?
    [itex]\frac{c^{2}}{9}[/itex] = 1 - x[itex]^{2}[/itex] - [itex]\frac{y^{2}}{4}[/itex]

    c = [itex]\sqrt{\frac{1 - x^{2} - \frac{y^{2}}{4}}{9}}[/itex]

    I don't know where i would go from there.


    The equation my instructor gave us for a ellipsoid is:
    [itex]\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}}[/itex] = d
    That's where I got c=3
     
  7. Oct 10, 2011 #6

    Mark44

    Staff: Mentor

    I'm not sure, but I don't think that c as used in this problem is related to the c in your ellipsoid equation. I believe they are using in the equation z = c to mean a plane that is parallel to the x-y plane, that's all.
     
  8. Oct 10, 2011 #7
    I think you're right Mark. I feel kind of silly for making that assumption. lol.

    The book says the answer is:

    [itex]\frac{2\pi(9 - c^{2})}{9}[/itex]

    I honestly don't see how it got that answer.

    Thanks for all the help so far!
     
  9. Oct 10, 2011 #8

    Mark44

    Staff: Mentor

    You're looking for the area of the elliptical cross-section that is cut by the plane z = c.

    Follow the directions that SammyS gave back in post #3, substituting c for z in the equation to get

    x2 + y2/4 = <something with c in it>

    Now divide both sides by what you have on the right side to get this equation into the standard form for an ellipse. You will then be able to pick out a and b, which you can use in your area formula for an ellipse.
     
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