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Double integral under a plane

  • Thread starter Bill333
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  • #1
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Homework Statement


Hi ! :) I'm having some difficulties with the question below, in which there are numerous steps and I am unsure in which part(/s!) I have gone wrong.

The question is as below; you must via integration calculate the shaded volume of a perfect cylinder of radius R and height h. The question wants you to do it as an integration under a plane I believe, so I have attempted to do so.
scan0006.jpg

Homework Equations




The Attempt at a Solution


Equation of a plane can be described by 3 points, which I have chosen as (0,R,h),(0,0,0) and (1,0,0). From this i have two vectors, A (0,-R,-h) and B(1,-R, -h) for which I have done the cross product to find the equation of a plane:
-hy +Rz= 0
I then integrate in polar coordinates over over the surface, which I believe is ∫∫z(x,y) dA = (h/R)∫∫r2sinθ dr dθ with the limits being 0≤ r ≤ R and 0 ≤ θ ≤π, which gives me the answer of 2R2h/3. I am dubious of this answer, as looking at the symmetry of the container I would assume it was (πR2/4)h.

Any clues towards where I went wrong would be highly appreciated!
Thanks in advance :)
Edit: Changed 'shaded area' to 'shaded volume'
 
Last edited:

Answers and Replies

  • #2
BvU
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Hi,
Is Dan's share a volume fraction or an area ?
 
  • #3
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Hi! :)
A volume fraction of the cylinder, apologies; the shaded volume, not the shaded area!
 
  • #4
LCKurtz
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I think your work and your answer are correct. The plane does not divide the left half of the cylinder into equal parts. If you calculate the volume of the half cylinder above the plane you will get ##\frac{R^2h\pi}{2}- \frac{2hR^2}{3}##.
 
Last edited:
  • #5
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Ah, thank you very much LCKurtz!
 

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