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Using the double integral to find the volume bounded betwee two solids

  1. Apr 24, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the volume of the solid T enclosed above by the sphere x^2+ y^2 + z^2 = 2 and below by the parabloid x^2 + y^2 = z

    2. Relevant equations

    The double integral. Possiblly polar coordinates (x = r*cos(theta) y = r*sin(theta)). z = f(x,y)

    3. The attempt at a solution

    I'm not really sure how to start this one since we have two equations with z. I know we want to find a function to represent z. I was thinking of using the parabloid, so that z = x^2 + y^2 = f(x,y). But I'm not sure of how the sphere comes into play. I also know that the z value is between 0 and root 2 (according to the sphere. And that x and y are bounded by the parabloid. I've tried looking in my calculus for help, but they only use one function with one z in it, not two.

    Also, I am supposed to be using a double integral to solve this. Thank you guys in advance for the help!
  2. jcsd
  3. Apr 24, 2009 #2


    Staff: Mentor

    Your integrand in a double integral is going to be the height (a difference of z-values) between the paraboloid (at the bottom) and the upper hemisphere of the sphere (at the top. To get the z-value for the sphere, solve for z in the equation x^2 + y^2 + z^2 = 2, and keep in mind that you want the upper half of the sphere.

    This difference in z-values is going to vary, depending on where x and y are. Hopefully you have drawn a sketch of the two surfaces so you can get an idea of what the solid looks like whose volume you're calculating, as well as where the two surfaces intersect, which will affect your limits of integration.

    I would probably do this as a polar integral because of the presence of the x^2 and y^2 terms in both surfaces; that is, because of the symmetry of both surfaces.
  4. Apr 24, 2009 #3
    I have drawn the sketch already, and I can see I'm dealing with the upper half of the sphere only. Solving for the in the sphere gives z = sqrt(2 - x^2 - y^2), which will be my integrand. I also know that the sphere and the parabloid intersect at the points along the curve sqrt(2 -x^2 - y^2) - x^2 - y^2 = 0. I am still a little confused about how I find the limits of integration
  5. Apr 24, 2009 #4
    I'm sorry, there's a typo in the one above. The integrand will be sqrt(2 -x^2 - y^2) - x^2 - y^2. I am still unsure of what the limits of integration will be
  6. Apr 24, 2009 #5


    Staff: Mentor

    Whether you do the integration in rectangular coordinates or polar, you're going to have to find where the paraboloid intersects the hemisphere. (It does so in a circle.) Your integration will be much simpler if done using polar coordinates, as the integrand will be sqrt(2 - r^2) - r^2. For limits of integration, theta is easy-- 0 to 2pi. r is a little harder, as it ranges from 0 to whatever its value is at the circle where the two surfaces intersect.
  7. Apr 24, 2009 #6
    Wouldn't the integrand be r*(sqrt(2-r^2) - r^2) when we go from rectangular to polar coordinates?

    As to where the two objects intersect, I know it occurs when x^2 + y^2 = sqrt(2 - x^2 - y^2). I'm not really sure how to solve for that. I know it means that r is equal to the square root of the right side, so the highest value of r is the quartic root of (2 - x^2 - y^2). Putting this into polar coordinates, we get quartic root of (2-r^2).

    I'm not sure if that's what I'm supposed to be doing to find the points of intersection.
  8. Apr 24, 2009 #7
    If I transform to polar coordinates, I get r^2 = sqrt(2 - r^2). Solving for this, I get r = 1, which would make the limits of integration for r is 0 to 1.
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