# Thread: [SOLVED] Volume of Revolution Question

1. ## [SOLVED] Volume of Revolution Question

Question:
Th region $R$ which is bounded by the curve $y = \frac{2}{x + 1}$, the $x-axis$, and the lines $x = 1$ and $x = 5$. Show that the volume of the solid formed when $R$ is rotated completely about the x-axis is $\frac{4}{3}\pi$

Attempt:

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

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

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

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

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

Why did I get the answer negative? Where did I go wrong?

2. Hello,

Actually, I think that the answer has to be :

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

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

$\pi$ or not, I don't know

3. integrating 2/(x+1)^2 gives you -2/(x+1)

ok i found the mistake >.<

you squared the function of y wrongly

the integration should be

integration from 1 to 5 of the function 2^2/(x+1)^2