# [SOLVED] Volume of Revolution Question

• Apr 25th 2008, 03:36 AM
looi76
[SOLVED] Volume of Revolution Question
Question:
Th region $\displaystyle R$ which is bounded by the curve $\displaystyle y = \frac{2}{x + 1}$, the $\displaystyle x-axis$, and the lines $\displaystyle x = 1$ and $\displaystyle x = 5$. Show that the volume of the solid formed when $\displaystyle R$ is rotated completely about the x-axis is $\displaystyle \frac{4}{3}\pi$

Attempt:

$\displaystyle y = \frac{2}{x+1}$

$\displaystyle \pi\int^5_1\left(\frac{2}{(x + 1)^2}\right)^2dx$

$\displaystyle \Rightarrow \pi\left[\frac{4}{(x+1)}\right]^5_1$

$\displaystyle \Rightarrow \frac{2}{3}\pi - 2\pi$

$\displaystyle = -\frac{4}{3}\pi$

Why did I get the answer negative? Where did I go wrong?
• Apr 25th 2008, 03:38 AM
Moo
Hello,

Actually, I think that the answer has to be :

$\displaystyle \int_1^5 \int_{\color{red}0}^{\frac{2}{x+1}} dy dx$

As you're bounded with the x-axis, that is to say y=0, you have to put it in the integral..

$\displaystyle \pi$ or not, I don't know :)
• Apr 25th 2008, 03:39 AM
integrating 2/(x+1)^2 gives you -2/(x+1)