Volume of the smaller part of a sliced cylinder

[songoku]

Homework Statement

A solid cylinder of radius r and height h is cut by a plane which passes through a diameter of the base and intersects the cylinder's top in just one point, cutting the cylinder into two parts. Calculate the volume of the smaller part

Relevant Equations

Integration

I imagine the shape will be like this:

and I need to find the volume of the shaded part. I am planning to use: $V=\int A(y) dy$

I tried to take the cross-sectional area A(y) to be triangle:

and the base of the cylinder to be:

So it means that the base of the triangle will fulfill the equation of circle $x^2+y^2=r^2$ and the base will be $x=\sqrt{r^2-y^2}$.

But I can't find the equation for the height of triangle. I imagine that the height will satisfy maybe equation of tilted parabola and I don't know how to find it.

Is my approach correct? Or maybe there is better method to do this question?

Thanks

 

[Orodruin]

What is the equation of the plane? What does this mean for when $x = \sqrt{R^2 - y^2}$?

That being said, I would have used the circle segment slices instead.

Edit: Actually, your slicing does result in nicer integrals. Keep doing it, just keep the above in mind. (Luckily, the result is the same)

 

[songoku]

What is the equation of the plane? What does this mean for when $x = \sqrt{R^2 - y^2}$?

I am sorry, which plane do you mean?

 

[Orodruin]

I am sorry, which plane do you mean?

The only plane mentioned in the problem statement.

Homework Statement:: A solid cylinder of radius r and height h is cut by a plane which passes through a diameter of the base and intersects the cylinder's top in just one point, cutting the cylinder into two parts. Calculate the volume of the smaller part

 

[songoku]

The only plane mentioned in the problem statement.

I think I get your hint and I got the volume to be $\frac{2}{3}r^2h$

Is this correct? Thanks

 

[Orodruin]

I think I get your hint and I got the volume to be $\frac{2}{3}r^2h$

Is this correct? Thanks

It is the same result as I obtained in both ways.

 

[songoku]

That being said, I would have used the circle segment slices instead.

For practice, I tried to use your method but I am confused about the slice. Do you take the slice as semicircle (half of the base area) then stacking it in upwards direction (integrate it from 0 to h)?

Thanks

 

[Orodruin]

For practice, I tried to use your method but I am confused about the slice. Do you take the slice as semicircle (half of the base area) then stacking it in upwards direction (integrate it from 0 to h)?

Thanks

No, it is not a semi-circle, it is a circular segment. I expressed the area of this as a function of $z$ and integrated with respect to $z$.

 

[anuttarasammyak]

Area of cross section of the body with the plane z = c is
\int_{rc/h}^r 2\sqrt{r^2-x^2} dx
Volume of the body is
\int_0^h dz\int_{rz/h}^r 2\sqrt{r^2-x^2} dx=2hr^2 \int_0^1 dz \int_0^{cos^{-1}z} sin^2\theta d\theta

[EDIT]
Let $\cos^{-1} z =t$
=hr^2 \int_0^1 dz [\theta-\frac{\sin 2\theta}{2}]_0^t
=hr^2 \int_0^{\pi/2} \sin t (t-\frac{\sin 2t}{2})dt
=hr^2 \{[-t \cos t]_0^{\pi/2}+ \int_0^{\pi/2} \cos t dt - \int_0^1 sin^2t d(\sin t)\}=\frac{2hr^2 }{3}

 

[songoku]

Thank you very much for the help Orodruin and anuttarasammyak

 

[Orodruin]

Area of cross section of the body with the plane z = c is
\int_{rc/h}^r 2\sqrt{r^2-x^2} dx
Volume of the body is
\int_0^h dz\int_{rz/h}^r 2\sqrt{r^2-x^2} dx=2hr^2 \int_0^1 dz \int_0^{cos^{-1}z} sin^2\theta d\theta

As mentioned in #2 though, using the triangular slicing of the OP results in significantly easier integrals.