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# Part 2 of shell method with 2 functions of y

Created by Sal Khan.

## Want to join the conversation?

• I know I am being nitpicky here, but isn't x=(y-1)^2 a relation, and not a function?
(1 vote)
• It is a function of y, but not of x.
At this level of study, you need to specify what it is a function of, especially when you have equations with more than two variables.
• Where can I find Part 1 of this video? Somehow I'm just overlooking it or am not able to find it.
• I did the search on Khan Academy, but I cannot find Part 1 of "shell method with 2 functions of y". I can only find "shell method with 2 functions of x". So, I would assume 1 of the following 2 situations:
1- Part 1 is, in deed, "shell method with 2 functions of x"
2- Part 1 of "shell method with 2 functions of y" may have been obsoleted and eliminate from the program, but noone ever bother to do the error correction in the title.
Anyone knows better the answer, please speak up. Otherwise, we should bring this up to the KA Team for error correction in order to eliminate the confusion for the learners. Thank you.
2-
(1 vote)
• Where is the video for part one?
• I've used disc method for this problem and after trying it multiple times I'm consistently coming up with 104pi/3.
(1 vote)
• The disc method is not applicable to this problem because we are now dealing with a shape defined by two function curves rotated around a separate axis: one for the outer boundary and one for the inner boundary. The disc method only deals with a single function curve used as an outer boundary rotated around a straight axis and assumes the shape contains all of the space between function and the axis.
• If the question is "y=4-x, y=1/2, y=2-x/3, and rotated about the x axis" then how can I find the r in order to calculate area?
(1 vote)
• shouldn't it be ((y-1)^2 -(y+1)) instead of what Sal did?
(1 vote)
• I don't think it really matters, since the difference is supposed to be a positive value in this case.
I'm not completely sure if you can do it that way, but it seems reasonable.
I tested the integral function from this video on my calculator with both (y+1)-(y-1)² and (y-1)² -(y+1), and both of them gave me the same absolute value --- (63·π)/2.