May 1st, 2017, 07:02 PM  #1 
Senior Member Joined: Apr 2008 Posts: 194 Thanks: 3  volume of revolution
A region is bounded by the curve y=x^3+x+1, x=1 and the xaxis. 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=1x and the height is h=x^3+x+1. Then, I integrate 2*pi*(1x)(x^3+x+1) from the xintercept on the negative side of the xaxis to 1. Now, I have a problem. I don't know how to find the xintercept. The exact value of the volume is 43*pi/30. Can someone explain how to determine the xintercept? Thanks a lot. 
May 1st, 2017, 07:41 PM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 3,034 Thanks: 1621 
something's not right ... using a calculator, $x^3+x+1 = 0$ at $x \approx 0.6823278038$ let $a = 0.6823278038$ $\displaystyle 2\pi \int_a^1 (1x)(x^3+x+1) \, dx \approx 7.598$ $\dfrac{43\pi}{30} \approx 4.503$ edit ... note that $\displaystyle \int_0^1 (1x)(x^3+x+1) \, dx = \dfrac{43\pi}{30}$, which would be the region in quadrant 1 between the cubic curve, the xaxis, and $x=1$. Last edited by skeeter; May 1st, 2017 at 08:18 PM. 
May 2nd, 2017, 02:06 PM  #3 
Member Joined: Feb 2015 From: Southwest Posts: 96 Thanks: 24 
I got a different answer for the integral from a to 1, but it is still off from the desired answer.

May 2nd, 2017, 05:20 PM  #4 
Math Team Joined: Jul 2011 From: Texas Posts: 3,034 Thanks: 1621 
Wolfram & my TI84 say about 7.6 ...

May 2nd, 2017, 09:55 PM  #5 
Member Joined: Feb 2015 From: Southwest Posts: 96 Thanks: 24 
Well crap, I put a zero after the decimal.


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