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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?
     
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  3. Nov 17, 2009 #2

    HallsofIvy

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    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]
    to
    [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]
    to
    [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: 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

    HallsofIvy

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

    HallsofIvy

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    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?
     
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