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: Volume of solid bounded by paraboloid and plane.

  1. Nov 17, 2009 #1
    1. The problem statement, all variables and given/known data

    Hi. I'm asked to find the volume of the solid bounded by the paraboloid

    4z=x^2 + y^2 and the plane z=4

    I have drawn the graph in 3D but I'm unsure of how to set up the integral. Also, how does one decide to use double integrals/triple integrals when finding volume?
  2. jcsd
  3. Nov 17, 2009 #2


    User Avatar
    Science Advisor

    You could use both! But since [itex]V= \int\int\int dV= \int\int\int dzdydx[/itex] but since the integral of "dz" is just z, if the boundaries can be written as z= f(x,y) and z= g(x,y), then that triple integral just reduces to the double integral [itex]\int\int\int (f(x,y)- g(x,y))dydx[/itex]

    Here the upper boundary is just z= 4 and the lower boundary is [itex]z= (1/4)(x^2+ y^2)[/itex], You could integrate
    [tex]\int\int\int_{z= (1/4)(x^2+y^2)}^4 dzdydx[/itex]
    or just write it as the double integral [itex]\int\int ((1/4)(x^2+y^2)- 4)dydx[/itex].

    Now, the limits of integration in x and y: When you graphed it you probably saw that the paraboloid and plane intersect where z= 4 and [itex]4(4)= 16= x^2+ y^2[/itex] which, projected to the xy-plane is the circle [itex]x^2+ y^2= 16[/itex] and the entire figure is inside that cylinder. So the limits of integration for x and y are given by that circle. One way to cover that circle is to take x from -4 to 4 and, for each x, y from
    [tex]-\sqrt{16- x^2}[/tex]
    [tex]\sqrt{16- x^2}[/tex]
    Or, because of symmetry, y from -4 to 4 and, for each y, x from
    [tex]-\sqrt{16- y^2}[/tex]
    [tex]\sqrt{16- y^2}[/tex].

    Perhaps simplest, because of the circular symmetry, is to use polar coordinates with r going from 0 to 4 and [itex]\theta[/itex] from 0 to [itex]2\pi[/itex]. That would be the same as setting up the entire in cylindrical coordinates: the boundaries would have equations z= 4 and [itex]4z= r^2[/itex] in cylindrical coordinates.
    Last edited by a moderator: Nov 17, 2009
  4. Nov 17, 2009 #3
    Thanks for the reply. If I wanted to set it up using polar coordinates (I haven't covered cylindrical coordinates yet) I would take r between 0 and 4, theta between 0 and 2pi, but over what function would I integrate?
  5. Nov 17, 2009 #4
    I can't seem to evaluate the integral using the "dydx" method, so I'm thinking this problem needs to be done using polar coordinates.
  6. Nov 17, 2009 #5


    User Avatar
    Science Advisor

    I believe I answered that in the second paragraph of my response.
  7. Nov 17, 2009 #6


    User Avatar
    Science Advisor

    Sure you can.
    [tex]\int_{x=-1}^1\int{y= -\sqrt{16-x^2}}^\sqrt{16-x^2} (4- (1/4)(x^2+ y^2)dydx[/itex]

    The first integral will give, of course, [itex]4y- (1/4)(x^2y+ (1/3)y^3)[/itex] evaluated between
    [itex]-\sqrt{16- x^2}[/itex] and [itex]\sqrt{16-x^2}[/itex] or [itex]8\sqrt{1-x^2}+ 2(x^2\sqrt{1- x^2}+ (1/3)(1- x^2)^{3/2}[/itex] and that looks like a candidate for a trig substitution like x= sin(t).

    But yes, it will be much simpler in polar coordinates.
  8. Nov 18, 2009 #7
    Well, I'm not sure what part of your post corresponds to the second paragraph >.<

    So does this integral look correct?

    Double integral from 0 to 4 and from 0 to 2pi

    (4-(x^2+y^2)/4)*r d(theta)dr = (4 - r/4)*r d(theta)dr
  9. Nov 19, 2009 #8
    Could I get a confirmation that this is indeed correct?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook