# Volume using integration ... Partial fraction decomposition

• Sep 19th 2013, 09:32 PM
kyliealana
Volume using integration ... Partial fraction decomposition
Consider the region R bounded by y = 1 / (x^2 + 3x + 2) , x = 0, x = 1 and the x-axis.

Find the volume of the solid formed by rotating R about the x-axis.

I need help with this problem.
I believe I am supposed to take the integral from 0 to 1 of pi(1 / (x^2 + 3x + 2))dx
I use partial fraction decomposition and get ...

1 = A(x+3)(x-1)^2 + B(x-1)^2 + C(x+3)^2(x-1) + D(x+3)^2

From substituting x = 1, D = 1/16
and x = -3, B = 1/16

I need help finding A and C. and I am also wondering if my reasoning is correct so far.
• Sep 19th 2013, 09:45 PM
Prove It
Re: Volume using integration ... Partial fraction decomposition
No, you actually need to evaluate \displaystyle \begin{align*} \int_0^1{ \pi \, y^2 \, dx} = \int_0^1{ \pi \left( \frac{1}{x^2 + 3x + 2} \right) ^2 \, dx} = \pi \int_0^1{ \frac{1}{\left( x + 2 \right) ^2 \left( x + 1 \right) ^2 } \, dx} \end{align*}

So your partial fraction decomposition will actually need to be \displaystyle \begin{align*} \frac{A}{x + 2} + \frac{B}{(x + 2)^2} + \frac{C}{x + 1} + \frac{D}{(x + 1)^2} \equiv \frac{1}{\left( x + 2 \right) ^2 \left( x + 1 \right) ^2} \end{align*}
• Sep 19th 2013, 11:51 PM
kyliealana
Re: Volume using integration ... Partial fraction decomposition
A = 4, B = -1, C = -3, D = 1 ?

I obtained 1 = x^3(A+C) + x^2(4A + 5C) + x(5A + 8C + 2) + (2A + 4C + 3)