1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Volume under curve z=1-x^2-2y^2

  1. Apr 12, 2012 #1
    1. The problem statement, all variables and given/known data
    z=1-x^2-2y^2
    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

    Mark44

    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

    Mark44

    Staff: Mentor

    I agree.

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

    $$4\int_{x = 0}^1\int_{y = 0}^{.25\sqrt{1 - x^2}}1 - x^2 - y^2~dy~dx$$
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook