What is the volume of a perfect cylinder under a plane?

In summary, the conversation is about a question involving the calculation of the shaded volume of a perfect cylinder using integration under a plane. The student presents their attempt at a solution, including the use of polar coordinates and the equation of a plane. They are unsure if their answer is correct and seek clarification from others. Another user confirms that the student's work and answer are correct, and explains why the symmetry of the cylinder may have caused confusion.
  • #1
Bill333
7
0

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:
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  • #2
Hi,
Is Dan's share a volume fraction or an area ?
 
  • #3
Hi! :)
A volume fraction of the cylinder, apologies; the shaded volume, not the shaded area!
 
  • #4
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:
  • Like
Likes Bill333
  • #5
Ah, thank you very much LCKurtz!
 

Related to What is the volume of a perfect cylinder under a plane?

1. What is a double integral under a plane?

A double integral under a plane is a type of mathematical calculation that involves finding the volume of a three-dimensional shape that lies beneath a flat plane. This can be visualized as finding the area of a shape on a coordinate plane, but in three dimensions.

2. How is a double integral under a plane calculated?

To calculate a double integral under a plane, the first step is to set up the integral using the given limits of integration and the function of the plane. Then, the integral is solved using techniques such as Fubini's theorem or integration by parts.

3. What is the significance of finding a double integral under a plane?

Finding a double integral under a plane is useful in many applications of mathematics and science. It can be used to find the volume of a solid object, the center of mass of an irregular shape, or the average value of a function over a given region.

4. Can a double integral under a plane have negative values?

Yes, a double integral under a plane can have negative values. This can occur if the function of the plane dips below the x-y plane, resulting in a volume with a negative value. However, the absolute value of the integral will still represent the total volume under the plane.

5. Are there any real-world applications of double integrals under a plane?

Double integrals under a plane have many real-world applications, particularly in physics, engineering, and economics. For example, they can be used to calculate the volume of a solid object, the work done by a force, or the expected value of a stock portfolio.

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