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May 1st, 2017, 07:02 PM   #1
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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.
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May 1st, 2017, 07:41 PM   #2
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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.
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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
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Well crap, I put a zero after the decimal.
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