# Cylindrical Shells

## Homework Statement

Use the method of cylindrical shells to find the volume generated
by rotating the region bounded by the given curves about the
specified axis. Sketch the region and a typical shell.

y = 4x - x^2, y = 3; about x = 1

?

## The Attempt at a Solution

int(2*pi*(1-x)*(3-4*x+x^2), x = 0 .. 1) = 11pi/6

This is the only problem on the book that I can't really understand. The correct answer is 8pi/3 according on the odd answers on the back page.

Increasing the upper limit to 2 gives the correct answer. But why? Isn't it I'll just integrate until 1 so I can get a y = 3. Why 2?

## Answers and Replies

Mark44
Mentor
Your limits of integration are wrong. The region over which you are integrating is [1, 3], not [0, 1].
I get 8pi/3 with this integral:
$$-2\pi \int_1^3 (x - 1)(x^2 - 4x + 3)dx$$

I think the mistake you are making is in the formula that you are integrating. It seems like you were thinking washers while using the shell equation.
The general equation to use for the shell method is

2$$\pi$$$$\int$$ R dx(dx can change depending on which variable you are integrating with respect to, but in this case it would be dx).

Where R is the distance from the axis of rotation. In this case, the R isn't 3-f(x), just f(x).

Mark44
Mentor
I think the mistake you are making is in the formula that you are integrating. It seems like you were thinking washers while using the shell equation.
Sarah's integrand was correct. Her only problem was she had the wrong limits of integration.
The general equation to use for the shell method is

2$$\pi$$$$\int$$ R dx(dx can change depending on which variable you are integrating with respect to, but in this case it would be dx).
No, this is not correct. The volume of a cylindrical shell is 2pi * radius * height * thickness. Your formula completely omits the height of the shell.
Where R is the distance from the axis of rotation. In this case, the R isn't 3-f(x), just f(x).

why would the limits of integration be from 1 to 3 if you are integrating with respect to the x axis?

nevermind, i've got it! sorry for my faulty answer and thank you for correcting my mistake.

Mark44
Mentor
No problem. We all make mistakes at least once in a while...

oh... now i noticed my problem.

4x - x^2 = 3
0 = x^2 - 4x + 3
0 = (x-3)(x-1)

x = 1 and 3

Tiredness really made my brain crazy.