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AP®︎ Calculus AB (2017 edition)
Course: AP®︎ Calculus AB (2017 edition) > Unit 11
Lesson 8: Volume: shell method (optional)- Shell method for rotating around vertical line
- Evaluating integral for shell method example
- Shell method for rotating around horizontal line
- Shell method with two functions of x
- Calculating integral with shell method
- Shell method with two functions of y
- Part 2 of shell method with 2 functions of y
- Shell method worksheet
- Shell method
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Shell method worksheet
Problem 1
Problem 2
Problem 3
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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?(13 votes)
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.(40 votes)
- 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?(3 votes)
To elaborate on finding the intercept point, you can set the equations equal to one another and solve for zero.(4 votes)
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.)(2 votes)
Yes there is symmetry.
Had the question been rotating about the axis x=0, then your tactic would have worked
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.(2 votes)
- 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.(2 votes)
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.(2 votes)
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)- The second problem's answer is wrong! plz fix(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 .(0 votes)
