Results 1 to 7 of 7

Thread: Solids of Revolution Question

  1. #1
    Newbie
    Joined
    Feb 2011
    Posts
    12

    Solids of Revolution Question

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

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

    So I don't know how they've reached the answer given.
    Any hints?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    I don't think that you have enough bounds on your region...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2011
    Posts
    12
    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 $\displaystyle \frac{7\Pi}{2}\ +\ \frac{\Pi\ ln\ 4}{2}$. My answer is equivalent to $\displaystyle \frac{7\Pi}{2}$, so I just can't see how they got the second part.

    Anyone else have any ideas?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Sorry I misread what you posted, I see now that the $\displaystyle \displaystyle y$ axis is also a bound.

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

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

    $\displaystyle \displaystyle = \frac{e^{2x}}{2} + 2e^x + x + C$.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Feb 2011
    Posts
    12
    Thank you for your reply.

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

    Getting closer, but I still don't understand how they got:
    $\displaystyle \frac{\Pi}{2}(7\ +\ ln\ 4)$
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Basic logarithm law: $\displaystyle \displaystyle n\log{x} = \log{(x^n)}$...
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Feb 2011
    Posts
    12
    Quote Originally Posted by Prove It View Post
    Basic logarithm law: $\displaystyle \displaystyle n\log{x} = \log{(x^n)}$...
    Thanks very much for your reply.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Solids of revolution question
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Aug 21st 2011, 12:35 PM
  2. Solids of Revolution
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Apr 20th 2010, 08:59 PM
  3. solids of revolution/integration question
    Posted in the Calculus Forum
    Replies: 5
    Last Post: Jan 4th 2010, 06:15 PM
  4. Solids of Revolution
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Feb 13th 2009, 11:42 AM
  5. solids of revolution
    Posted in the Calculus Forum
    Replies: 11
    Last Post: Jun 9th 2007, 01:30 PM

/mathhelpforum @mathhelpforum