Region of y=1/(x^2+4x+5)

[confused_07]

This is my last homework question for the class!!!! And it's a good one...

R is the region that lies between the curve y= 1/(x^2+4x+5) and the x-axis from x= -3 to x= -1. Find:

A) the area of R
B) the volume of the solid generated by revolving R around the y-axis
C) the volume of the solid generated by revolving R around the x-axis

I found the first one:

A= int [a,b] f[x] dx
= int [a=-3,b=-1] 1/(x^2+4x+5) dx
= int [a=-3,b=-1] 1/[(x+2)^2+1] dx
= [arctan(x+2)] [a=-3,b=-1]
= 45-(-45) = 90


I am stuck on the other two since you have to square f[x] as part of the equation. When I do, I come out with:

1/(x^4+8x^3+26x^2+40x+25) which I can factor to:

1/[x^2(x^2+8x+26)+25]

From there, I don't know what to do? Would I do the method of partial fractions?

 

[galactus]

B) the volume of the solid generated by revolving R around the y-axis

Shells:

\(\displaystyle \L\\-2{\pi}\int_{-3}^{-1}\frac{x}{x^{2}+4x+5}dx\)

When a quadratic like this won't factor, a good thing to do is complete the square.

\(\displaystyle \L\\-2{\pi}\int_{-3}^{-1}\frac{x}{(x+2)^{2}+1}dx\)

Now, let u=x+2 and you have an arctan and an ln in your future.

C) the volume of the solid generated by revolving R around the x-axis

Washers:

\(\displaystyle \L\\{\pi}\int_{-3}^{-1}\frac{1}{(x^{2}+4x+5)^{2}}dx\)