Cylindrical shell method to compute volume of a napkin ring

Question

Use cylindrical shells to compute the volume of a napkin ring of height 3h created by drilling a hole with radius r through the center of a sphere of radius R and express the answer in terms of h.

 

I've been going around for a while on this one. I thought I was going to use a quarter circle rotated, but that doesn't take into account the portion of the sphere that is lobbed off at both poles when a cylinder is drilled through it. Because of that, I don't know how to establish a relationship with the height so that the answer can be in terms of h.

Answer 1

Using Shell Method,

 

  Volume V= 2*(2π∫_r^R x*√(R^2-x²)dx,  where  √(R²-x²) is height of the semi-circle

 

  let u= R²-x², du= -2xdx,  xdx= -du/2

  Then, ∫ x*sqrt(R²-x²)dx =(-1/2)∫ (u)^1/2du = (-1/2)(2/3)u^3/2 = -(1/3)u^3/2

 

  Therefore, V = 2*(2π)(-1/3)(R²-x²)^3/2|_r^R = -4π/3(( R²-R²)^3/2 - (R²-r²)^3/2) )= 4π/3(R²-r²)^3/2 ,

     where R²-r² = (3h/2)², R is a radius of outer circle, which is hypotenuse  ⁄ , r is a radius of inner circle _ , and 3h/2 is a half of the height of cylinder |,  of a right triangle ⁄_{_}|

   

                    

      ∴     V = 4π/3((3h/2)²)^3/2 = (4π/3)(3h/2)³ = 9πh³/2

 

Answer 2

Hi Judith, after wrestling with this one for a while, I think I have an answer - it's difficult to explain some calculus problems without drawings, but I'll try to make it clear!

 

As we look at the napkin ring from the side, we can imagine a right triangle formed against the inside wall of the hole: The hypotenuse, R, which is the radius of the sphere, the horizontal bottom leg with a distance of r, which is the radius of the cylindrical hole, and the vertical leg of 3h/2 (because we are only looking at half of the total height of the napkin ring). 

      /|

     / |

    /  |

R /   | 3h

  /    |  2

 /     |

/      |

   r

 

The Pythagorean Theorem tells us that

 

R² = r² + (3h/2)²             or   (3h/2)² = R² - r²

 

We aren't going to use this just yet, but we will need it later.

 

Now let's try to figure out the volume. We can pick an arbitrary cylindrical shell, much like a really thin tin can without a top or bottom, and the dimensions of which will again be based on a right triangle from the center of the sphere. The shell that is right against the inside whole would be associated with exactly the right triangle we just looked at, but let's look at a generic shell, not just that inside one. To do that, we'll define some new variables. Any particular shell would be a certain horizontal distance away from the center of the sphere (I'll call that x), and it would have a height to it. On the triangle below, I'll call the vertical distance y, but the total height of the shell would be 2y, because the shell would be both above and below the center of the sphere. The hypotenuse of the triangle would still be R, the radius of the sphere, for any shell that exists.

 

       /|

      / |

     /  |

    /   |

R /    | y

  /     |

 /      |

/       |

   x

 

The volume of any one of these shells would be equal to the circumference of the circle it makes (as if you were looking down at the tin can from above), times the height, times the thickness:

 

dV = (2πx) * (2y) * dx             = 4πxy dx

            circum.*height*thickness

 

In order to integrate, we want to put the equation all in terms of just one variable, so let's come up with a way to express y in terms of x by looking at the triangle above. Again, from the Pythagorean theorem,

 

R² = x² + y²

 

y² = R² - x²

 

y = √(R² - x²)

 

Now that we have y in terms of x, we will substitute it into the volume equation we had before:

 

dV = 4 π x √(R² - x²) dx

 

We can now integrate with respect to x, from the smallest horizontal x distance (r) to the largest (R)

 

Volume = 4 π ∫_r^R x √(R² - x²) dx

 

Let's use substitution to make the integral easier.

 

u = (R² - x²)          and       du = -2x

                                         dx 

So dx = - du

               2x

 

This makes our integral: (I am leaving off the bounds of the integral, since we are not going to evaluate it in terms of u, anyway - I'll put them back once we actually do the integral and put everything back in terms of x)

 

Volume = 4π ∫ w √(u) (-du)        and after combining/cancelling:

                             2w

 

Volume = -2π ∫ √(u) du

 

Now we do the actual integral, remembering that a square root is the same thing as having an exponent of (1/2):

 

Volume = -2π * 2 * u^(3/2)

                        3

 

Now combine terms, and replace u with the x expression:

 

Volume = -4 * π * (R² - x²)^(3/2), evaluated from (r to R) 

                3

 

Volume = [-4 * π * (R² - R²)^(3/2)] - [-4 * π * (R² - r²)^(3/2)]

                 3                                   3

 

The left term, with (R² - R²), will simply be zero, leaving:

 

Volume = 4 * π * (R² - r²)^(3/2)

               3

 

Now let's go back to that first triangle we drew, which we can now use to make a substitution, because we know that 

 

(3h/2)² = R² - r²

 

This makes the equation:

 

Volume = 4 * π * ((3h)²)^(3/2)

               3             2

 

We can simplify the exponents:

 

Volume = 4 * π * (3h)³
               3           2

 

Volume = 4 * π * 27 * h³

               3          8

 

Volume = 9 π h³

               2