My Math Forum volume of revolution

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 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 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.
 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 $a = -0.6823278038$ $\displaystyle 2\pi \int_a^1 (1-x)(x^3+x+1) \, dx \approx 7.598$ $\dfrac{43\pi}{30} \approx 4.503$ 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 by skeeter; May 1st, 2017 at 08:18 PM.
May 2nd, 2017, 02:06 PM   #3
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I got a different answer for the integral from a to 1, but it is still off from the desired answer.
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May 2nd, 2017, 05:20 PM   #4
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Wolfram & my TI-84 say about 7.6 ...
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 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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