volume of revolution

Apr 2008
194
3
A region is bounded by the curve y=x^3+x+1, x=1 and the x-axis. Find the exact volume of the solid by revolving the region about the line x=1.

my solution

I use the cylindrical method to solve the problem. The radius is r=1-x and the height is h=x^3+x+1.

Then, I integrate 2*pi*(1-x)(x^3+x+1) from the x-intercept on the negative side of the x-axis to 1.

Now, I have a problem. I don't know how to find the x-intercept.

The exact value of the volume is 43*pi/30. Can someone explain how to determine the x-intercept? Thanks a lot.
 

skeeter

Math Team
Jul 2011
3,363
1,854
Texas
something's not right ...

using a calculator, $x^3+x+1 = 0$ at $x \approx -0.6823278038$

let $a = -0.6823278038$

$\displaystyle 2\pi \int_a^1 (1-x)(x^3+x+1) \, dx \approx 7.598$

$\dfrac{43\pi}{30} \approx 4.503$

edit ...

note that $\displaystyle \int_0^1 (1-x)(x^3+x+1) \, dx = \dfrac{43\pi}{30}$, which would be the region in quadrant 1 between the cubic curve, the x-axis, and $x=1$.
 
Last edited:
Feb 2015
96
24
Southwest
I got a different answer for the integral from a to 1, but it is still off from the desired answer.
 

Attachments

Feb 2015
96
24
Southwest
Well crap, I put a zero after the decimal.