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Volumes of Irregular Shapes by Integration

  1. Sep 18, 2014 #1
    Hi all,

    Is it possible to determine the volume of a shape where, in the x and y dimensions, the shape is described by an equation, and then its elevation is described by another equation?

    An example would be a parabola in the x-y plane whose elevation is based on another parabolic function. For example lets say [tex]y=x^2+4[/tex] from [itex]x=-2[/itex] to [itex]x=2[/itex] with an elevation determined by a circle of radius 4 (so zero elevation at the apex and the x-axis). Meaning [tex]z=\sqrt{16-y^2}[/tex] (independent of x).

    Can I get this volume by integration, or would I just need to apply some kind of finite element approach? Thanks in advance; most all of the volume integrals I could find online or in a text are just volumes of revolution.
     
  2. jcsd
  3. Sep 18, 2014 #2
    yes, in order to do this it requires that the bounds of integration be functions instead of numbers.

    [itex]\int_{x_{0}}^{x_{1}}\int_{y_{0}(x)}^{y_{1}(x)}z(x,y) dydx[/itex]

    here y1(x) is greater than y0(x) throughout the interval where x0<x<x1
    look up "double integrals" for more.

    also, given your problem it would be far easier to use a change of coordinates(most likely polar coordinates)
     
  4. Sep 18, 2014 #3
    Ah yes. I knew I was missing something (it requires that the bounds of integration be functions instead of numbers). I was having trouble making that mental leap. And I will look into the polar coordinates. I hadn't thought of those for a few years :). Thanks!
     
  5. Sep 25, 2014 #4

    HallsofIvy

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    The problem here is that this is not a solid at all. With just the information that "[itex]y= x^2+ 4[/itex]", (x, y) is restricted to that parabola and adding z (height) just gives a "wall" which has area, not volume. Perhaps you meant to add another boundary and have x and y inside the bounded region? Say "y between [itex]y= x^2+ 4[/itex] and [itex]y= 12- x^2[/itex]. Those two curves intersect at (-2, 8) and (2, 8). With height [itex] \sqrt{16- y^2}[/itex], the volume is given by [tex]\int_{x= -2}^2\int_{y= x^2+ 4}^{12- x^2}\sqrt{16- y^2} dy dx[/tex]

     
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