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Homework Help: Volume under curve z=1-x^2-2y^2

  1. Apr 12, 2012 #1
    1. The problem statement, all variables and given/known data
    find volume under curve
    bounded by the xy plane.
    is the answer sheet wrong? (see below)
    why am i struggling so much with this??!?!?! how do i do it?

    2. Relevant equations
    according to other answer sheet, it is pi/sqrt 2

    3. The attempt at a solution
    i did this, its bound by z=0 and it should be the same in all the upper quadrants so i tried too work out one quadrant and went from there.
    4*integral(0 to 0.25) (integral(0 to 1)(1-x^2-2y^2) dx)dy =0.625
    although im wondering if it should be
    4*integral(0 to 1) (integral(0 to 0.25sqrt(1-x^2)(1-x^2-2y^2) dy)dx =0.564504
  2. jcsd
  3. Apr 12, 2012 #2


    Staff: Mentor

    The limits on the inner integral aren't right.
    This setup looks right, but I haven't carried out the integration, so can't confirm that your answer is correct.
  4. Apr 12, 2012 #3
    It would help after 150 posts if people made the effort to put at least some parts of it into latex.
  5. Apr 12, 2012 #4


    Staff: Mentor

    I agree.

    brandy, here is your second integral using LaTeX. Right-click the integral to see my LaTeX script.

    $$4\int_{x = 0}^1\int_{y = 0}^{.25\sqrt{1 - x^2}}1 - x^2 - y^2~dy~dx$$
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