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# Shell method worksheet

## Problem 1

A region is enclosed by the x-axis and the curve y, equals, 3, x, minus, x, squared.
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, equals, 1, minus, x, squared.
What is the volume of the solid generated when this region is rotated around the line x, equals, 1?

## Problem 3

A region is enclosed by the line y, equals, x and the parabola x, equals, left parenthesis, y, minus, 2, right parenthesis, squared.
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.
• The second problem's answer is wrong! plz fix
• 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.
• 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.
(1 vote)
• For problem two, I do not understand how the given solution can be correct: I did everything identically through finding the antiderivatives but when evaluating for negative 1, the signs for 1/2 and 1/4 seem swapped. Can someone either confirm this or explain to me what I am missing? thanks
(1 vote)
• Why does happen when you choose the variable you get the same amount as you started with but multiplied
(1 vote)
• What is flux? How can be calculate flux using shell method ?
(1 vote)
• The bounds will usually be where two curves intersect.

In #1, we have y=3x-x^2 and the x-axis. The x-axis is actually the line y=0, so to find where y=3x-x^2 hits the line y=0, we let 0=3x-x^2. Then, 0=x(3-x), and our solutions are x=0 and x=3. These are the bounds of integration.

For #2, we have y=1-x^2 and the x-axis (y=0) again. So let 0=1-x^2. Then 1=x^2, and our solutions are x=±1, so our bounds of integration are -1 and 1.

For #3, we have x=y and x=(y-2)^2, so we'll let y=(y-2)^2. Expanding the right side of the equation gives y=y^2 -4y + 4, then subtracting y from both sides gives 0=y^2 -5y + 4. We can factor this to 0=(y-1)(y-4), so our solutions (and therefore our bounds of integration) are y= 1 and y=4

hope this helps ♥
(1 vote)
• what happens when we take vertical infinitesimal strip instead of horizontal (which we did in problem no. 3)?
Does it leads to disk? If yes, does its answer is same .