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Volume of a solid

  1. Jan 29, 2004 #1
    The base of a solid is the region bounded by the graphs of x=y^2 and x=4. "Each cross section is perpendicular to the x-axis is a triangle of altitude 2." Find the volume of the solid.

    That was how it was worded, I'm guessing it meant Each cross section perpendicular to the x-axis is a triangle of altitude 2?

    Assuming that's what it says. How are these problems tackled? The addition of the quoted sentence confuses me a tad bit on what direction to head.
  2. jcsd
  3. Jan 29, 2004 #2
    So after looking at it some more, I got:

    Int{0 to 4} Sqrt(3)/4 * (4 - Sqrt(x))^2 dx ?

    Hrmm.. ok, I just noticed I didn't use the altitude...
    Last edited: Jan 29, 2004
  4. Jan 30, 2004 #3


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    Good guess! Did the problem really have that extraneous first "is"?

    Since the area of a triangle is (1/2)h*b and h= 2, you only need to calculate b. The cross section is perpendicular to the x-axis so the base is the y distance. y, for specific x, ranges from x2 up to 4 so the distance is b= 4-x2. That is, the area of such a triangle is (1/2)(4-x2)(2)= 4- x2. Imagining each cross section as an infinitesmally this slab, of thickness dx (since the thickness, perpendicular to the plane, is in the x-direction), the "volume" of each slab is (4- x2)dx.

    Putting all of the "slabs" together, the total volume is

    I have absolutely no idea where you got all those square roots!
  5. Feb 1, 2004 #4
    shouldn't the volume of a solid

    shouldn't it be composed as a triple integral since you are dealing with a solid? or did I not read the problem correctly?
  6. Feb 1, 2004 #5
    HenryF...might I suggest

    this helped me a lot when I was in calc...

    making a graphical representation (even if it is 3 dimetions) can help a lot.
  7. Feb 2, 2004 #6


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    It could be but it is not necessary. Since you were told that every cross section is "a triangle of altitude 2", you can use the area formula to find the area of each cross section. If the cross section had been a more general figure, you might have had to use a double integral to find that area.
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