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Math Help - Solids of Revolution Question

  1. #1
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    Solids of Revolution Question

    The Problem
    The finite region bounded by the curve y\ =\ e^x\ +\ 1, the y-axis, and the line x\ =\ ln\ 2, is rotated through 360 degrees about the x-axis. Show that the volume of the solid formed is \frac{\Pi}{2}(7\ +\ ln\ 4).

    Attempt
    Volume = Integral(\Pi\ y^2\ dx), with limits 0\ and\ ln\ 2.
    This becomes: Volume = \Pi\ Integral((e^x\ +\ 1)^2), with limits 0\ and\ ln\ 2.
    (We can take pi outside the integral and multiply the result by it later).
    Integrating term by term:
    Integral((e^x\ +\ 1)^2)\ =\ \frac{e^{2x}}{2}\ +\ 2e^x + 1.
    Substituting ln\ 2:
    \frac{e^{2\ ln\ 2}}{2}\ +\ 2e^{ln\ 2}\ +\ 1\ =\ 7.
    Substituting 0:
    \frac{e^{2(0)}}{2}\ +\ 2e^{0}\ +\ 1\ =\ 3\ \frac{1}{2}.
    Subtracting:
    7\ -\ 3\ \frac{1}{2}\ =\ 3\ \frac{1}{2}.
    So my answer is:
    3\ \frac{1}{2}\ \Pi.

    So I don't know how they've reached the answer given.
    Any hints?
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  2. #2
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    I don't think that you have enough bounds on your region...
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  3. #3
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    Quote Originally Posted by Prove It View Post
    I don't think that you have enough bounds on your region...
    I'm not exactly sure what you mean by this. This is from a textbook, so do you mean that the information given does not give enough bounds to determine the volume of the region, or do you mean that the bounds of 0 and ln 2 that I have used are wrong?

    Also, I found that if you multiply out the answer given, you get \frac{7\Pi}{2}\ +\ \frac{\Pi\ ln\ 4}{2}. My answer is equivalent to \frac{7\Pi}{2}, so I just can't see how they got the second part.

    Anyone else have any ideas?
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  4. #4
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    Sorry I misread what you posted, I see now that the \displaystyle y axis is also a bound.

    Your integral \displaystyle \int_0^{\ln{2}}{\pi y^2\,dx} = \pi \int_0^{\ln{2}}{(e^x + 1)^2\,dx} is correct, however you evaluated this integral incorrectly.

    \displaystyle \int{(e^x + 1)^2\,dx} = \int{e^{2x} + 2e^x + 1\,dx}

    \displaystyle = \frac{e^{2x}}{2} + 2e^x + x + C.
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  5. #5
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    Thank you for your reply.

    Substituting ln 2:
    \frac{e^{2\ ln\ 2}}{2}\ +\ 2e^{ln\ 2}\ +\ ln\ 2\ =\ 6\ +\ ln\ 2
    Substituting 0:
    \frac{e^{2(0)}}{2}\ +\ 2e^0\ +\ 0\ =\ 2\ \frac{1}{2}
    Subtracting:
    6\ +\ ln\ 2\ -\ 2\ \frac{1}{2}\ =\ 3\ \frac{1}{2}\ +\ ln\ 2\ =\ \frac{7}{2}\ +\ ln\ 2
    Multiplying by \Pi:
    \frac{7\Pi}{2}\ +\ \Pi\ ln\ 2
    Factorising:
    \frac{\Pi}{2}(7\ +\ 2\ ln\ 2).

    Getting closer, but I still don't understand how they got:
    \frac{\Pi}{2}(7\ +\ ln\ 4)
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    Basic logarithm law: \displaystyle n\log{x} = \log{(x^n)}...
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  7. #7
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    Quote Originally Posted by Prove It View Post
    Basic logarithm law: \displaystyle n\log{x} = \log{(x^n)}...
    Thanks very much for your reply.
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