# Thread: How to calculate the volume of partial cylinder?

1. ## How to calculate the volume of partial cylinder?

Hello from Finland!
I have a problem how to calculate the volume and the mass centroid of a partial cylinder. Please look at the image below. I have listed the given information there. Variable for a is: a=0...2R

Ive tried to solve it, but I dont know how to do it... All help is appreciated! Thanks!
-Janne

2. ## Solution when a=R

I found that the solution for a=R is:
V = 2/3*R^2*b
c = R-3/16*PI*R

3. Hyvää on! Miten selvisi?

4. Originally Posted by TKHunny
Hyvää on! Miten selvisi?
Solution for a=R I found from a Finnish tablebook called "Rakentajain Kalenteri 2004".
But I need to also find solutions when a is not R...

5. I don't understand the drawing.

Is the height 'b' or something greater than 'b'?

Are we using Calculus or only Geometry?

6. ## Clarification for b

Originally Posted by TKHunny
I don't understand the drawing.

Is the height 'b' or something greater than 'b'?

Are we using Calculus or only Geometry?
Hi Tk

Height is b. I clarified the drawing below.

Basically Im hoping for an solution for V(olume) and mass (c)entroid in terms of b, a and R.

7. Very good. Unfirtunately, I am not at home and have only limited PC access. Perhaps another able source can help you with this before I get around to in on Thursday. Sorry.

8. Definitions and Orrientation:

I set it up so that:
1) The y-axis is parallel to the cut.
2) The x-axis is the perpendicular bisector of the cut.
3) The z-axis is parallel to 'b'.
4) The entire structure is in the four quadrants were z > 0.
5) The Origin of the x-y plane is at the center of the circle from the top view.

This has two side effects.
1) The centroid for the y-axis is obviously zero (0), due to symmetry.
2) The equation of the cut-plane is (z-0) = (-b/a)(x-R)
3) The centroid for the x-axis should be < 0. Intuitively clear.

The volume, then, would be:

$\displaystyle 2*\int_{-R}^{R-a}\int_{0}^{\sqrt{R^{2}-x^{2}}}\int_{0}^{b}\;dz\;dy\;dx + 2*\int_{R-a}^{R}\int_{0}^{\sqrt{R^{2}-x^{2}}}\int_{0}^{\frac{-b}{a}(x-R)}\;dz\;dy\;dx$

Let's see if you can get that far or if we need a different kind of solution.

9. Reality Check:

The Volume is easier than that, ceratinly for a = R, but we'll need this machinery for the Centroid.

For a = R

1) Uncut semi-circular region x < 0

$\displaystyle \frac{1}{2}\cdot \pi R^{2}b$

2) Cut region, x > 0

$\displaystyle \frac{1}{3}\cdot \frac{1}{2}\cdot \pi R^{2} b$

Sure enough, this matches your answer from the book.

10. Are you still with me?

In THIS case, since we have symmetry in 'y', we can ignore it and just work with z and x. This makes it rather simple, since the x-z cross section is a simple trapezoid.

$\displaystyle M = \frac{R+(R-a) + 2R}{2} \cdot b$

$\displaystyle M_{x} = \int_{-R}^{R-a}\int_{0}^{b}\;x\;dz\;dx + \int_{R-a}^{R}\int_{0}^{-\frac{b}{R-a}\cdot (x-R)}\;x\;dz\;dx$

Can you do $\displaystyle M_{z}$?

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