Calculus students: a required classic -- solid of revolution

🌟Support the channel🌟

🌟my other channels🌟

🌟My Links🌟

🌟How I make Thumbnails🌟

🌟Suggest a problem🌟

[neilgerace]

If r= 0, V = the volume of the sphere alone.

[TJStellmach]

When r=R/sqrt(2), the cylinder cuts out half the volume of the sphere.

[bradsword]

I guess for the satisfying values at the end:

If r=0, then you aren’t drilling a hole at all and the formula generates the volume of the sphere as 4/3*pi*r^3.

If r=R, then you are drilling away the entire sphere and the formula gives zero for the volume of the “null solid of revolution.”

[RexxSchneider]

As this question has been asked twice in the comments, I thought I'd present the general solution. We are given the half-height (h) of the cylinder and asked to find the residual volume.
From the calculation that Michael Penn did to find the upper limit of integration, it should be obvious that is the same value as the half-height of the cylinder. So h^2 = R^2 - r^2
So the remaining volume = 4π/3 * (R^2 - r^2)^(3/2) = 4π/3 * h^3.
This shows that if h is fixed, the remaining volume does not depend on the values of R and r.
You can drill a large hole through a large sphere and the remaining volume is exactly the same as drilling a very small hole through a small sphere, as long as the length of the hole is the same. In the limit as r → 0, the half-height of the "hole/cylinder" is equal to the radius of the smallest sphere, whose volume then matches the formula I derived above.

[TheRandomFool]

I used the shell method around the y-axis. The volume is then 4*pi* integral of x*sqrt(R^2-x^2) from r to R.

[txikitofandango]

Absolutely ideal board work. Clear and accurate and meaningful

[andrewmcintire]

You can calculate the surface area of this object as a function of R and r. You can actually get more surface area with the hole bored through it than the original sphere but only up to a certain hole size. As a fun fact, you can find the size of the hole that maximizes the surface area and the result is quite nice.

[txikitofandango]

Definitely R = 5 and r = 4, which yield a volume of 36π (I think)

[JonathanMandrake]

I had to do calculation for exactly two types of revolutions:
The first was the one formed by z=x^p for some natural number p.
We had to find out for which p the volume of this revolution was finite but the surface infinite, a generalisation of Gabriels horn.
The second was the volume of a sphere, as an introduction into the use of spherical coordinates

[anon]

This was fun to solve and the result is amazing.
The result volume is the same volume as that of a sphere with radius equal to half the length of the cylinder ^_^

[stevenwilson]

Choose Pythagorean Triplets for R and r where R = hypotenuse (largest value) and r = either of the "legs". Obvious choices: 3-4-5 (R = 5, r = 3 or 4, if you choose 4, you'll remove the 3 on the bottom of the fraction). 5-12-13, R = 13, r = 12 or 5. 7-24-25, R = 25, r = 7 or 24. 8-15-17, choose R = 17, and r = 8 or 15 if you choose r = 8, you'll remove the 3 on the bottom of the fraction.

Finally 9-40-41, where R = 41, r = 40, you'll get a difference R^2 - r^2 = 9^2, and lose the 3 on the bottom.

[buxeessingh]

You can also assign how to represent the solid in cylindrical and spherical co-ordinates and then do the computation.

[misterdubity]

I did it as a double integral, maybe easier to visualize. 2*integral(theta=0 to 2pi) integral(x=r to R) sqrt(R^2-x^2)*x dx dtheta. same result

[dsacton]

Michael, a fun way to restate this problem is as follows: suppose you only know the half-height of the cylinder; no radii or any other dimensions? What is the volume then? How does this compare to the volume of a sphere?

[dragandraganov]

If we take R=5 and r=4, R^2-r^2=3^2 to the three halves, we get 27 and after multiplying by 4pi/3, the answer is 36pi which is very satisfying answer

[s]

If you take the cylindrical part as vertical the volume only depends on the height [because the height is sqrt(of R²-r²)] as shown before from Vasuce. (Napkin ring problem)

[alainbarnier]

Nice ! And with R=2 & r=1, it's V=4pi x sqr(3)... if my calculus is ok ?

[teeweezeven]

I think I saw a vsauce video about this!

"Hey vsauce, Michael (Penn) here"

[walifercosta]

For nice results we need R to be hyp for a pitagoric triangle similar whit the 3 4 5 type and r be one of the other sides, i don't speak english so i hope i write this correct kkkk
Obs: i don't test whit others potagorics triangles so maybe works too

[iabervon]

I saw the diagram and immediately thought I'd need to find the residues at the poles in the upper half-plane.