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Help with double integral - volume between 2 surfaces

  1. Apr 13, 2013 #1
    1. The problem statement, all variables and given/known data

    find volume of the solid bounded by the surfaces

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

    and x^2/4 -x +(Y^2)/2 = 0

    and the planes z = 0 and z = 1

    2. Relevant equations

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

    and x^2/4 -x +(Y^2)/2 = 0

    3. The attempt at a solution

    I think the first surface is an ellipsoid with it's highest point at z =1 and x = 0 and y = 0 and the second "surface" I have interpreted as a cylinder whose base is an ellipse centred at x = 2 and y =0. So it seems like there could be two solids here, the first would have an elliptical base of the z = 0 plane and the top would be the surface of the first equation above and the sides would be the sides of the cylinder. But it seems like I could also have another similar solid where the top is the z =1 plane and the base is the surface of the first equation.

    I think I must be a long way off track.
     
    Last edited by a moderator: Apr 13, 2013
  2. jcsd
  3. Apr 13, 2013 #2

    haruspex

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    Not an ellipsoid. A cone, of sorts?
    I agree the z=1 plane seems redundant.
     
  4. Apr 13, 2013 #3
    ok so we have now got it to the point where we are integrating from z goes from the xy plane up to the cone surface, and the area of integration is the elliptical cylinder.

    so in polar coordinates I think the integrand is 1 -r/2

    but I am not sure what the limits of integration are for r. The equation of the elliptical cylinder is ((x-2)/2)^2 + y^2/2 =1 so the ellipse doesn't have it's centre on the origin. does this mean the limits of integration for r is not 0 to 1?
     
  5. Apr 13, 2013 #4

    haruspex

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    I think it's a bit clearer if you take a slice at some z value. What does the region of integration look like?
     
  6. Apr 13, 2013 #5
    since the base we are integrating over is an ellipse with a = 2 and b = root2, we have converted to elliptic coordinates x = 2r cos(θ) and y = sqrt(2)r cos(θ)

    and integrating the top function from my original post. when I convert it to elliptic coordinates it reduces to
    (1 - r/2) r dr dθ

    but now I'm not sure what the limits of integration for r should be.

    I thought maybe I need to transform both funtions so the volume I am finding is centred at the origin?

    Or is there a better way?
     
  7. Apr 13, 2013 #6

    haruspex

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    But what does the whole region look like at height z? The elliptical cylinder only provides one part of the boundary.
     
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