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Math Help - Volume of Revolution

  1. #1
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    Volume of Revolution

    Find the volume of revolution when the area between y=x^2 + 1 and y=x+3 is rotated about the x-axis.

    (x+3)^2 = x^2 + 6x + 9
    (x^2+1)^2 = x^4 +2x^2 + 1

    So subtract those and I got:

    -x^4 - x^2 + 6x + 8

    Using S for integral as I don't know the code for it.

    V = (pie)S(-1 to 5) (-x^4 - x^2 + 6x + 8)
    Going to jump into the substitution with the anti-deriv.
    =(pie)(-(2)^5 / 5 - (2)^3 / 3 + 3(2)^2 + 8(2)) - (-(-1)^5 / 5 - (-1)^3 / 3 + 3(-1)^2 + 8(-1))

    Got a final answer of 119/5 but my friend got a different answer of 77/5. What did I do wrong if mine is incorrect
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  2. #2
    Super Member Random Variable's Avatar
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    The limits are wrong. It's from -1 to 2.
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  3. #3
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    Quote Originally Posted by Random Variable View Post
    The limits are wrong. It's from -1 to 2.
    Ooops wrote that wrong...the work on my paper uses 2
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  4. #4
    Super Member Random Variable's Avatar
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    I'm getting \frac {117}{5} \pi

     \pi \int^{2}_{-1} \Big( (x+3)^2 - (x^{2}+1) \big) \ dx

     = \pi \int^{2}_{-1} \Big(-x^{4}-x^{2}+6x+8 \Big) \ dx

     = \pi \Big( -\frac {1}{5}x^{5} - \frac {1}{3} x^{3} + 3x^{2} + 8x \Big) \Big|^{2}_{-1}

     \pi \bigg(\Big( -\frac {32}{5} - \frac {8}{3} + 12 + 16 \Big) - \Big( \frac {1}{5} + \frac {1}{3} + 3 - 8 \Big) \bigg)

     = \frac {117}{5} \pi
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  5. #5
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    The radii are incorrect. You are rotating the region around the x-axis. The closest point to the x-axis of the function x^2+1 is (0,1) and the closest point of the function x+3 is (-1,2). Therefore, the radii should be:

    Rin = (x^2 + 2)^2
    Rout = (x + 5)^2

    So, your integral should be something like this:

    V = \pi \int -x^4-3x^2+10x+21 dx, from -1 to 2

    I got: \frac{312\pi}{5}

    Edit: Can someone verify my answer, please.
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