Cylindrical shells

  • Thread starter Benny
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Hi I'm stuck on an integration problem where I need to use the method of cylindrical shells to calculate a volume.

Q. Using the method of cylindrical shells, find the volume generated when the area bounded by the curve y = x^2 - 3 and the line y = 2x is revolved about the line x = 7.

The main thing that I'm having trouble with is setting up the integral. I think it's due to a limited conceptual understanding of what's going on with questions of this type. I started by drawing a quick sketch and finding the x values of intersection, x = +/- sqrt(3).

I drew a cylinder, with the vertical axis x = 7 going through its centre, around the region.

Volume(cylinder) = (circumference)(height)(thickness) = CHT

[tex]
C = 2\pi r = 2\pi \left( {7 - x} \right)
[/tex]

[tex]
H = 2x - \left( {x^2 - 3} \right) = - x^2 + 2x + 3
[/tex]

T = dx

[tex]
V = 2\pi \int\limits_{ - \sqrt 3 }^{\sqrt 3 } {\left( {7 - x} \right)\left( { - x^2 + 3x + 3} \right)dx}
[/tex]

That's what I came up with. I can't check to see if my answer is correct because I don't have the solution to the question. Any help with this question would be good thanks.
 

Answers and Replies

  • #2
HallsofIvy
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Well, the first thing you've done wrong is miscalculate the x values of the intersection of the curves. They are not [tex]\plusminus\sqrt{2}[/tex]!
Show how you got that value and we may be able to point out your mistake.
Other than that, it looks like you have the right integral.
 
  • #3
saltydog
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So what'd ya get Benny? To find the intersection of two curves f(x) and g(x), just set them equal to each other right? So set those two equal, solve for x and you have the limits of integration. I made a plot.
 

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  • #4
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Hmm...x^2 - 3 = 2x => (x-3)(x+1) = 0 => x = 3 or -1. I don't know how I got +/- sqrt(3) the first time around. Thanks for pointing that out.

So continuing from before(the quadratic should have a 2x term instead of 3x in my previous post).

[tex]
V = 2\pi \int\limits_{ - 1}^3 {\left( {7 - x} \right)} \left( { - x^2 + 2x + 3} \right)dx
[/tex]

[tex]
= 2\pi \int\limits_{ - 1}^3 {\left( {x - 7} \right)\left( {x^2 - 2x - 3} \right)dx}
[/tex]

[tex]
= 2\pi \int\limits_{ - 1}^3 {\left( {x^3 - 9x^2 + 11x + 21} \right)dx}
[/tex]

[tex]
= 2\pi \left[ {\frac{{x^4 }}{4} - 3x^3 + \frac{{11x^2 }}{2} + 21x} \right]_{ - 1}^3
[/tex]

[tex]
= 2\pi \left( {\left( {\frac{{81}}{4} - 81 + \frac{{99}}{2} + 63} \right) - \left( {\frac{1}{4} + 3 + \frac{{11}}{2} - 21} \right)} \right)
[/tex]

[tex]
\mathop = \limits^{calculator} 128\pi
[/tex]

Edit: BTW saltydog, which program do you use to draw those sketch graphs?
 
  • #5
saltydog
Science Advisor
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Benny said:
[tex]
\mathop = \limits^{calculator} 128\pi
[/tex]

Edit: BTW saltydog, which program do you use to draw those sketch graphs?
That's what I got. I use Mathematica. Try and check it out. Your school should have it. You know, Mathematica is also good to back-substitute solutions into ODEs to verify the solutions. Good for lots of other things too.:smile:
 
  • #6
OlderDan
Science Advisor
Homework Helper
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Benny said:
[tex]
\mathop = \limits^{calculator} 128\pi
[/tex]

Edit: BTW saltydog, which program do you use to draw those sketch graphs?
saltydog said:
That's what I got. I use Mathematica. Try and check it out. Your school should have it. You know, Mathematica is also good to back-substitute solutions into ODEs to verify the solutions. Good for lots of other things too.:smile:
Some Mathematica functions are avilable online at

http://www.quickmath.com/

Click on equations and advanced equation plotting page
 

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