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### Course: Calculus, all content (2017 edition)>Unit 6

Lesson 11: Shell method

# Shell method worksheet

## Problem 1

A region is enclosed by the $x$-axis and the curve $y=3x-{x}^{2}$.
What is the volume of the solid generated when this region is rotated around the $y$-axis?

## Problem 2

A region is enclosed by the $x$-axis and the curve $y=1-{x}^{2}$.
What is the volume of the solid generated when this region is rotated around the line $x=1$?

## Problem 3

A region is enclosed by the line $y=x$ and the parabola $x=\left(y-2{\right)}^{2}$.
What is the volume of the solid generated when this region is rotated around the $x$-axis?

## Want to join the conversation?

• May I ask how to recognize the methods used in different test questions? How to recognize when to use the Washer Method, when to use the Disk method, and the Shell method?
• Very great question! I believe Sal should make a video explaining when to use these. Here is my opinion on it:
Disc method - When the solid of revolution is a single function rotated about a line. If the solid of revolution is solid throughout, and can be sliced into many thin circles stacked on top of each other, the disc method is typically easiest. For example, y = x² rotated about the y-axis, or y = √(x) + 1 rotated about y = 1.
Washer method - A generalization of the disc method, for two functions rotated about a line. If the solid of revolution has a hollow inner section, and it can be sliced into many thin "washers", or thin donuts, the washer method can be used. For example, the region bound by x² and √(x) rotated around x = 2, or the region bound by the x-axis and 1 - x² rotated about y = 2.
Shell method: Can be used for all functions, but typically for functions that are hard to be expressed explicitly. Functions can be sliced into thin cylindrical shells, like a piece of paper wrapped into a circle, that stack into each other. For example, y = x(x - 1)³(x + 5) from [-5, 0] rotated about the y-axis (hopefully a good example, because the washer method would be difficult to use).
Hope that I helped, and correct me if I'm wrong.
• When determining the volume of a shell, I'm still confused on how you are getting the bounds? Specifically in problem 3. How did you determine what a and b was in problem 3?
• To elaborate on finding the intercept point, you can set the equations equal to one another and solve for zero.
• For problem 2, I saw that the function was even and the interval was symmetrical, and decided to factor out 2 and evaluate the integral on the interval (0,1). The answer I got was 5pi/3. Why didn't this give the correct answer?
(I checked my work; I hope it's not an arithmetic mistake.)
• Yes there is symmetry.
but since the axis of rotation is x=1, the part of the parabola between -1 and 0 sweeps out a larger area/volume than the part between 0 and 1.

Thought experiment:
Just using a 2d analog, what is the area swept out by the line segment from 0 to 1?
It is a circle with area pi*r^2 = pi*1^2 = pi.

Now what is the area of the circle swept out from -1 to 1?
It is a circle with area pi*2^2 = 4pi.

If you remove the area of the of the 0 to 1 circle from the -1 to 1 circle, you get 4pi - pi = 3p.
This shows that the area generated by rotating the line segment from 0 to 1 about the line x=1 is pi and the area swept out by rotating the line segment from -1 to 0 about the line x=1 is 3pi.
The areas are not symmetrical! and neither are the volumes.
• I am having a hard time figuring why x has a larger value for x=y than for x=(y-2)^2.
What makes it larger?
(1 vote)
• I pluged in a value in the interval to see which one was larger. I used 3. (3-2)^2=1, while 3=3. Therefore, y>(y-2)^2.
• For question 3, I do not understand why the height is y-(y-2)^2 not (y-2)^2-y as (y-2)^2 is the larger number.
(1 vote)
• For question 3, try tilting your head 90 degrees to the right, and notice how y is larger than (y - 2)^2. We are integrating with respect to y (notice the dy).

Edit: Good job on your streak! We have the exact same streak of 122 days right now.