1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Finding a volume in space

  1. May 18, 2009 #1
    Hello, I need to find the volume between the plane z=1 and z=10-x*x-4*y*y.

    I have tried using polar coordinates:
    first integrate from z=1 to z=10-x*x-4*y*y
    The I integrate in the plane z=1, but here I need to integrate over an ellipse, how do I do this? When I wrote the expression to polar coordinates I got an expression for r that I couldnt integrate.

    I have also tried integrating over the ellipse using cartesian coordinates, but I got stuck aswell, can someone pleas help?
  2. jcsd
  3. May 18, 2009 #2


    User Avatar
    Homework Helper

    Could you show the actual function you have to integrate over? I suspect it has the form [itex]\sqrt{9-4y^2}[/itex]? If so use the substitution [itex]u=\frac{3}{2} \sin \theta[/itex]. Still integrating like this isn't very pretty. A more elegant method and by far the easiest to integrate is to notice that this is a paraboloid with elliptical cross section. You can then slice the paraboloid parallel to the x-y plane in a lot of slices and then integrate from z=1 to 10. The only thing you have to do is find an expression for the semi major axis and the semi minor axis as a function of z and use that the area of an ellipse is given by pi a b, with a the semi major axis and b the semi minor axis.
    Last edited: May 18, 2009
  4. May 18, 2009 #3


    User Avatar
    Science Advisor

    No, finding the volume bounded by z= 1 and [itex]z= 10- x^2- 4y^2[/itex] involves integrating the difference in z values, [itex](10- x^2- 4y^2)- 1= 9- x^2- y^2[/tex] over the circle [itex]x^2+ 4y^2= 9[/itex], where the two surfaces cross.

    Now the limits of integration, in xy-coordinates will involve something like [itex]y= \frac{1}{2}\sqrt{9- x^2}[/itex].

    Hyper, you might consider using parameters r and [itex]\theta[/itex] such that [itex]x= r cos(\theta)[/itex] and [itex]y= 4r sin(\theta)[/itex], not standard polar coordinates, with r going from 0 to 1 and [itex]\theta[/itex] from 0 to [itex]2\pi[/itex]. Do you know how to find the "differential of area" in r and [itex]\theta[/itex] using the Jacobian?
  5. May 18, 2009 #4


    User Avatar
    Homework Helper

    Why wouldn't slices be applicable, it yields the same answer?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook