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lij Group Title

1a) suppose we want to integrate a funciton f(x,y) over the ellipse x^2/a^2 + y^2/b^2=1 (where a is not equal to b are positive constants). Suggest a suitable change of variables, and find the Jacobian of this transformation. (use a modified version of polar coordinates) b) Using the answer to (a), find the volume of the region enclosed beneath the elliptic paraboloid z = 12-x^2-3y^2 and above the xy-plane.

  • 2 years ago
  • 2 years ago

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  1. lij Group Title
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    I have no idea where to start. I tried to convert the coordinates into polar coordinates (x=rcos(theta), y=rsin(theta). I don't know even where to start...

    • 2 years ago
  2. Jemurray3 Group Title
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    A much better parameterization would be x = a cos (theta), y = b* sin(theta)

    • 2 years ago
  3. lij Group Title
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    okay so, I have donethat and got the determinant of the jacobian =abr how do i do part 2?....

    • 2 years ago
  4. Jemurray3 Group Title
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    There should be no r involved, I'm not sure what you did there.

    • 2 years ago
  5. Jemurray3 Group Title
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    Okay, I read this a little more closely and my thought is this: The first question is, to me, poorly posed because your domain of integration is a 1-dimensional curve. You only need one coordinate to specify where you are on that ellipse (theta). If you're trying to integrate over the domain enclosed WITHIN the ellipse, the situation changes. Now you need to consider your variable r. I would suggest a transformation similar to what you had before, but rather x = (r/a) cos(theta), y = (r/b) sin(theta)

    • 2 years ago
  6. Jemurray3 Group Title
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    I'm sorry, that should be (r*a) and (r*b), not (r/a) and (r/b)

    • 2 years ago
  7. lij Group Title
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    I think we figured it out...I'm not sure though...the assignment's not due till the end of the week, haha

    • 2 years ago
  8. lij Group Title
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    THANKS!

    • 2 years ago
  9. Cardioid Group Title
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    Rather than depending on the Internet to get help on the assignment, you can instead visit me during my office hours.

    • 2 years ago
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