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Homework Help: Find the volume

  1. Sep 13, 2006 #1

    tony873004

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    I’ve dropped my Calc II class because my Calc I class, which I took at the community college, never covered anti-derivatives, and my Calc II class started off assuming you already knew anti-derivatives and integration.

    So now I have the fun task of teaching myself anti-derivatives and integration so I can take Calc II again next semester. So I might as well attempt the problems given as homework to the Calc II class. But unlike the homework, I’m going to choose the odd numbered problems so I can check my answers.

    Q. Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. [tex]
    y = x^2 ,\,\,x = 1,\,\,y = 0;\,
    [/tex]about the x-axis.

    After drawing it, I came up with
    [tex]
    \begin{array}{l}
    v = \sum\limits_0^1 {\pi r^2 } \Delta x = \pi \sum\limits_0^1 {r^2 } \Delta x \\
    \\
    v = \pi \int\limits_0^1 {r^2 ,\,dx} \\
    \end{array}
    [/tex]

    Now here’s where my lack of anti-derivative skills hurt me. What do I do next? To get the anti-derivative of r squared, do I add 1 to the exponent and divide the whole thing by the new exponent? Should I get [tex]
    \pi \frac{{r^3 }}{3}
    [/tex]? If so, how do I apply this new formula to get an answer? My integral goes from 0 to 1. How do I get from here to the final answer of [tex]
    \pi /5
    [/tex]?
     
  2. jcsd
  3. Sep 13, 2006 #2
    Remember what an anti derivative is. This will help you check yourself.

    example)

    [tex] \int x^4 \,\, dx = \frac{1}{5}\,x^5 + C [/tex]

    Now differentiate it.

    [tex] \frac{d(\int x^4 \,\, dx)}{dx} = \frac{d(\frac{1}{5}\,x^5 + C)}{dx} = \frac{1}{5} \, \left( \frac{d(x^5)}{dx} + \frac{d(C)}{dx}\right) = \frac{1}{5} \, 5x^4 + 0 [/tex]

    .... hmmm I just noticed you put
    [tex]
    v = \pi \int\limits_0^1 {r^2 ,\,dx}
    [/tex]

    Did you mean for the [itex] r^2 [/itex]?
     
  4. Sep 13, 2006 #3

    tony873004

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    To get the volume, I'm adding up all the areas of the slices of the shape. They'll form disks with a thickness of delta x when rotated about the x-axis, so for the area I want to do pi r squared. Since radius of each slice would be y(x), I should have put x2 there instead of r2, especially since I put dx.
    [tex] v = \pi \int\limits_0^1 {x^2 ,\,dx} [/tex]
     
    Last edited: Sep 13, 2006
  5. Sep 13, 2006 #4
    Ok, well:

    [tex] \pi \int x^2 \, dx = \pi \frac{x^3}{3} [/itex]
    (setting the constant equal to 0)

    [tex] \pi \int_a^b x^2 \, dx = \pi \left( \frac{b^3}{3} - \frac{a^3}{3} \right) [/tex]
     
  6. Sep 13, 2006 #5

    tony873004

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    Or maybe I meant
    [tex] v = \pi \int\limits_0^1 {(r^2) ,\,dx} [/tex]
    give me a few minutes to think about this.
    edit ^^ my tex didn't do what I wanted it to do
     
  7. Sep 13, 2006 #6

    tony873004

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    [tex]
    \begin{array}{l}
    v = \pi \int\limits_0^1 {\left( {x^2 } \right)^2 \,,\,dx} \\
    v = \pi \int\limits_0^1 {x^4 \,,\,dx} \\
    v = \pi \frac{{x^5 }}{5} \\
    v = \frac{{\pi x^5 }}{5} \\
    \end{array}
    [/tex]

    So if I plug in 1 for x, I get the same answer as the back of the book. What would I have done if the integration went from 2 to 3 instead of 0 to 1? Do I subtract these numbers and plug it in the formula?
     
  8. Sep 14, 2006 #7
    Let's say you have a general function (that's integrable) [itex] f(x) [/itex]

    Let the antiderivative equal [itex] F(x) [/itex].

    Thus,

    [tex] F(x) = \int f(x) \, dx [/itex]

    [itex] F(x) [/itex] is called an indefinite integral.

    If you were to evaluate this integral (as a definite integral), you would have.

    [tex] \int_a^b f(x) \, dx =F(b) - F(a)[/tex]

    This is the first fundemental http://mathworld.wolfram.com/FirstFundamentalTheoremofCalculus.html" [Broken]of calculus.

    This answers your question in general.

    But to answer your question specifically.

    You asked, what if I integrate from 2 to 3 instead.

    [tex] \pi \int_2^3 x^4 \, dx [/tex]

    From the first fundemental theorem.
    [tex] \int_a^b f(x) \, dx =F(b) - F(a)[/tex]
    [tex] a = 2 [/tex]
    [tex] b = 3 [/tex]
    [tex] f(x) = x^4 [/tex]

    Now we have to find [itex] F(x) [/itex]
    [tex] F(x) = \int x^4 \, dx = \frac{x^5}{5} + C[/tex]

    Now plugging in a, b
    [tex] F(b) = F(3) = \frac{3^5}{5} + C [/tex]
    [tex] F(a) = F(2) = \frac{2^5}{5} + C [/tex]

    Finally [itex] F(b) - F(a) [/itex] is equal to:
    [tex] \left(\frac{3^5}{5} + C \right) - \left( \frac{2^5}{5} + C \right) [/tex]

    Notice that the constant is dropped.
    Also note that this was multiplied by [itex] \pi [/itex] !!!
     
    Last edited by a moderator: May 2, 2017
  9. Sep 14, 2006 #8
  10. Sep 14, 2006 #9

    J77

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    tony - you're going a bit wrong bringing the r into play.

    For solids of revolution, you just need the curves y=f(x).

    Sketch these curves first.

    You can then apply the formula:

    [tex]V=\pi\int_0^1f(x)^2dx[/tex]
     
  11. Sep 14, 2006 #10

    tony873004

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    Thank you FrogPad for your help and for the links.

    And Thanks J77. I figured that out that I needed to replace r with the function that gives me r, and it must be in terms of x.
     
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