# Thread: Solids of Revolution Question

1. ## 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:
$\displaystyle 7\ -\ 3\ \frac{1}{2}\ =\ 3\ \frac{1}{2}$.
$\displaystyle 3\ \frac{1}{2}\ \Pi$.

So I don't know how they've reached the answer given.
Any hints?

2. I don't think that you have enough bounds on your region...

3. Originally Posted by Prove It
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?

4. 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.

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

$\displaystyle \displaystyle = \frac{e^{2x}}{2} + 2e^x + x + C$.

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:
$\displaystyle \frac{\Pi}{2}(7\ +\ ln\ 4)$

6. Basic logarithm law: $\displaystyle \displaystyle n\log{x} = \log{(x^n)}$...

7. Originally Posted by Prove It
Basic logarithm law: $\displaystyle \displaystyle n\log{x} = \log{(x^n)}$...