Volume and Revolutions Calc Problem

Question

At Newniz Doughnuts they create their doughnuts in a different and unique way. Inspired by their calc class, they make doughnuts by spinning a circle though some dough 360 degrees around. When finished revolving the machine has a ring of dough which they cook into a doughnut. The diameter of the circle is 6 cm and the length of the stick attached to the revolving ring is 4 cm which is attached to the machine that revolves the dough.

One pound of mix makes 500 cm^3 of doughnut.

(1) What is the volume in cm^3 of one doughnut made with the machine. Use Wolfram Alpha to find the integral once you find it.
(2) How many pounds of mix (round to three decimal places) must be used to make 12 doughnuts?

Answer 1

I'll show you what might be a neat way to figure this out first, because I accidentally found the area of the circle instead of the volume of the torus when I first set up the integrals.

 

I'm going to say that the circle has a radius of r and is set R above the origin at point (0, r + R).

 

The equation for the circle would be

 

x² + (y - (R + r))² = r²

 

I'll solve this for y and write it as two different functions:

 

y_t = (R + r) + √(r² - x²)

 

y_l = (R + r) - √(r² - x²)

 

If you accidentally plop these into integrals, you can work out that the area of the circle is given by

 

A = 4∫[0, r]√(r² -x²)dx

 

But then you go, whoops, I already know the area of the circle without having to do the integral.

 

I really need to find the volume of the torus.

 

I do this by doing the disk method on y_t and y_l separately and subtracting the second volume from the first (draw a picture).

 

In the argument of those integrals you know you'll need to have πf²(x)dx to get the volume of the disks, where we replace f(x) with y_t and y_l alternatingly.

 

I'll take x between 0 and r, so I'll only be finding half of the volume at first:

 

v/2 = π∫[0, r]y_t²dx - π∫[0, r]y_l²dx

 

Put π on the other side for now and combine the integrals:

 

v/(2π) = ∫[0, r](y_t² - y_l²)dx

 

square your two ys and subtract separately before writing a bunch more messy integral stuff. Then

 

v/(2π) = 4(R + r)∫[0, r]√(r² -x²)dx

 

v/(2π) = (R + r)A

 

v = 2π(R + r)A

 

where A is the area of that little circle, πr²

 

v = 2π²r²(R + r)

 

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The nice thing about this is that that first mistake I made made it so we didn't have to actually compute the integral!

 

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That probably beats the point of the exercise, but you were going to plug it into something to do the integral for you anyway...

 

In this particular problem, r = 3 and R = 4

 

so the volume of the donut should be 

 

v = ?

 

Then the volume of 12 donuts is 12v and the pounds it takes to make this is what you get when you multiply 1/500 lb/cm³ by 12v. 

 

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If I misinterpreted the part about the stick being attached to the revolving ring, it could be that the stick goes all the way to the center of the circle instead of the outside edge (I did it thinking outside edge). Account for that if you have to. You'd use 4 for R + r instead of for R in the formula for volume, but that should be it.