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Homework Help: Integrals, Finding Volume definition

  1. Jan 23, 2008 #1
    [SOLVED] Integrals, Finding Volume definition

    1. The problem statement, all variables and given/known data
    I'm going through my textbook, working on some of the examples, and I just don't get what they're asking. I don't need this solved (The solution is right in the book), however I just don't grasp what they're asking me to do.

    "The base of the solid S is the region bounded by the parabola y = 9 - x^2 and the x-axis. Each cross-section of S perpendicular to the x axis is a square with one side in the xy-plane. Find volume V of S."

    So when I read this (And look at the picture, which is just a graph of y = 9 - x^2), I figure they're asking me to find the area under the graph.
    So, to find the integral of the function, take the antiderivative, then compute the values for the leftmost and rightmost points (-3 and 3 respectively) of the shape.

    But no. The solution does what I just said, only they squared (9-x^2) first, before taking the integral. Why?

    When they say perpendicular to the x axis, I figure they mean parallel to the y axis.
    But then they mention the xy-plane. If they need to specify the plane we're working in, then we must be using 3d, not 2d... in which, by perpendicular, do they mean along the z axis? And why the hell do we have a square in 3d anyway? What purpose is this?

    Someone help me untangle this mess please :(
  2. jcsd
  3. Jan 23, 2008 #2
    It would appear that you completely missed this section of your book:


    It's almost like a "trick"(though to call it a trick is to imply it's some type of gimmick when in fact it's a useful tool, but it gives off that vibe)to find the volume of an irregular solid using calculus 1 methods instead of resorting to what would, often, be a nasty triple integral

    Knowing what it's called, you can examine that link and your book to attempt to understand why it's squared and why they integrate like they do and why it gets the desired answer. If you can't, let us know. It's a classic place for people to start having trouble in calc

    Edit: Was there a factor of pi in the solution as well?

    Usually the problem is phrased in giving you a curve then telling you which axis it is rotated around to produce the solid. This information is given in a slightly different manner by telling you about the cross section and what axis it's perpendicular too.
    Last edited: Jan 23, 2008
  4. Jan 23, 2008 #3
    lol... the funny thing is, this example is on page 352.
    Solids of Revolution is taught on page 353.

    Regardless, I'll take a look. Thanks for the pointer.

    Edit: There was no pi in the answer, or anywhere in the solution.
  5. Jan 23, 2008 #4


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    Hmm, I can't quite see if this is a solid of revolution; I certainly don't see it. I understand it to be as follows. We need to find the volume of S such that the height function of S (the z value) is defined to be equals to y. Hence we get z=y=9 - x^2. The reason why the solution squared y= 9 - x^2 (bearing in mind that z=y), was because they were first calculating the area of a square tile of thickness dx lying on its edge on the xy plane, which make up S. Then the integral is taken over the interval (-3,3), to find the volume of S.

    EDIT: They are making use of the fact that the volume of a solid can be found by finding the cross-sectional area at one point, then multiplying it by dx to find dV, which is the volume of an infinitesimal cross-sectional piece. Then the integral is taken over the length of the solid, and by length I mean the axis which is perpendicular to the cross-sectional pieces each of thickness dx.
    Last edited: Jan 23, 2008
  6. Jan 23, 2008 #5
    The solution:
    Integral from -3 to 3 of "(9-x^2)^2" dx
    = Integral from -3 to 3 of "81 - 18x^2 + x^4" dx

    Antiderivative = 81x - 6x^3 + (x^3)/5
    Take the values of this from -3 to 3

    You get 1296/5.
  7. Jan 23, 2008 #6
    That's their introduction to it then

    I really need to draw it or see it but I'm at work and they frown on me filling pages with what's obviously not my job ^_^

    It must not be exactly a solid of revolution and I can't picture it in my head worth a darn. Try to understand what the solid itself looks like, and remember that area is always, in general, the area of the base times the height

    you're given the base of the solid

    Edit: Yah I was looking at it wrong, I totally didn't understand what was meant by square tile and stuff, and didn't realize the height was equal to y

    Like defennnder said, you're basically finding the area of a horizontal square "slice", then adding those areas up from bottom to top
    Last edited: Jan 23, 2008
  8. Jan 23, 2008 #7
    I sorta have an idea of what it'd look like... can't picture it quite yet, but I think I get the idea.

    Why would this mean we square the integral?
  9. Jan 23, 2008 #8


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    Well actually he's integrating (9-x^2)^2 now. But can you integrate x^2? If you can, just do the integration term by term.

    If you can't, then think of it this way, what polynomial function of x must I differentiate to get x^2?

    EDIT: The above was in reply to a post that was deleted Nevermind.
  10. Jan 23, 2008 #9


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    The integral is squared because the area of each cross-sectional square tile is yz, and y=z.
  11. Jan 23, 2008 #10
    http://chuwm2.tripod.com/revolution/solid4.gif [Broken]

    Now, this isn't a solid of revolution, but it's the same idea. You're finding the area of an infinitely thin "slice"(which in your case is a square as given by the problem)and summing those slices up, since the slices are infinitely thin, there are infinite slices, and the sum's an integral. The sum of those areas is the volume
    Last edited by a moderator: May 3, 2017
  12. Jan 23, 2008 #11
    So wait... the largest cross-sectional square would have an area of 36, right?
    Then the cross-sectional square when x=2 would have an area of 25, etc?

    Assuming this is right, then I believe I can visualize the shape.

    So, at any time, the height of this cross-sectional would be the same as the base, and the base is 9-x^2?

    So then the volume is the same as the area under the graph of 9-x^2, except now we multiply by height, which is also 9-x^2.... giving us (9-x^2)^2.

    Am I on the right track here?
  13. Jan 23, 2008 #12
    *finally drew it and gets the picture*

    I don't think it would've been immediately obvious to me if defennnder hadn't stated it already. Anyhoo, don't picture an integral as an area under the curve. That's useful, but not for these applications. Picture the integral as its more general definition of an infinite sum

    Now keep in mind that .gif I posted, it's actually more similar than I thought. In that picture, the solid was composed of a bunch of disks, I think you can see that.

    This solid is composed of a bunch of squares. The area of a square is a side^2. The problem basically states, and thanks to defennnder for easily understanding this, that the height of each square is equal to its function value, ie its y value

    you're given what y equals, and because of that statement, you're given what the height equals(you can call it z), and of course being a square they're the same, y=z=9-x^2

    The area of each square then is Asquare=y^2=z^2=(9-x^2)^2

    Now, volume in this case is the area of each slice times the thickness of each slice. We will make the thickness of each slice be infinitely small, dx(the .gif I sent may help)

    Therefore Asquare*dx=volume of each infinitely small slice. The integral of THAT, ie the infinite sum of an infinite number of infinitely small volumes = the total exact volume
  14. Jan 23, 2008 #13
    Yay, my education is saved! ^^;

    Thank-you all :P
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