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Math Help - Finding Revolving Volumes

  1. #1
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    Finding Revolving Volumes

    I've been able to calculate a lot of my homework involving finding the volume of a revolving function around either the x or y axis. However, this one seemed to pose a challenge for me.

    Find the volume of the solid formed by rotating the region enclosed by

    about the -axis.


    If anyone could provide an explanation and solution to this problem, it would be greatly appreciated. I'd really like to understand what this problem is asking for.

    Thanks!

    -C.C.
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by C.C. View Post
    I've been able to calculate a lot of my homework involving finding the volume of a revolving function around either the x or y axis. However, this one seemed to pose a challenge for me.

    Find the volume of the solid formed by rotating the region enclosed by

    about the -axis.


    If anyone could provide an explanation and solution to this problem, it would be greatly appreciated. I'd really like to understand what this problem is asking for.

    Thanks!

    -C.C.
    This is a standard problem using the shell method:

    V=2\pi\int_a^b xf(x)\,dx

    Plug in a=0, b=0.7 (the bounds) and f(x)=e^{4x}+4
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  3. #3
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    I think I'm having problems integrating:

    Integral of x(e^(4x)dx

    I used integration by parts, and I got:

    x((1/4)e^(4x)) minus the integral of (1/4)e^(4x)+4x dx

    And I found the antiderivative of the integral. I got:

    (1/16)e^(4x)+2x^2

    What did I do wrong?
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  4. #4
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by C.C. View Post
    I think I'm having problems integrating:

    Integral of x(e^(4x)dx

    I used integration by parts, and I got:

    x((1/4)e^(4x)) minus the integral of (1/4)e^(4x)+4x dx

    And I found the antiderivative of the integral. I got:

    (1/16)e^(4x)+2x^2

    What did I do wrong?
    u=x, dv=e^{4x}\,dx
    du=dx, v=\frac{1}{4}e^{4x}

    \frac{xe^{4x}}{4}-\frac{1}{4}\int e^{4x}\,dx

    ...
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  5. #5
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    I tried that integral. I got 30.67, and that was incorrect.
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  6. #6
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by C.C. View Post
    I tried that integral. I got 30.67, and that was incorrect.
    Sorry I should have been clearer.

    \int x(e^{4x}+4)\,dx=\int xe^{4x}+\int 4x\,dx

    I was only showing you how to integrate the first integral (on the right side).

    Taking what I had from before, we get that the whole thing is

    \frac{xe^{4x}}{4}-\frac{e^{4x}}{16}+2x^2\bigg|_0^{0.7}

    Plug in the bounds and you get 2.89. Multiply by 2\pi to get the final answer of 18.17.
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  7. #7
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    Oh, I understand much better! I've been really struggling with this one.

    Thanks. I'm so happy now!
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