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Part 2 of shell method with 2 functions of y

Created by Sal Khan.

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  • aqualine ultimate style avatar for user Joseph Adaoag
    I know I am being nitpicky here, but isn't x=(y-1)^2 a relation, and not a function?
    (1 vote)
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  • blobby green style avatar for user Samarth Goyal
    Where can I find Part 1 of this video? Somehow I'm just overlooking it or am not able to find it.
    (2 votes)
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    • blobby green style avatar for user Thien D Ho
      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)
  • blobby green style avatar for user Kyle Le Roux
    Where is the video for part one?
    (2 votes)
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  • leaf green style avatar for user Harrison
    I've used disc method for this problem and after trying it multiple times I'm consistently coming up with 104pi/3.
    (1 vote)
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    • leaf blue style avatar for user Kevin Jones
      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.
      (3 votes)
  • marcimus pink style avatar for user dwy0831
    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)
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  • leaf green style avatar for user dahria
    shouldn't it be ((y-1)^2 -(y+1)) instead of what Sal did?
    (1 vote)
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    • leaf green style avatar for user Peter Barke
      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.
      (2 votes)
  • piceratops sapling style avatar for user k.barbee95
    Where is part 1 of this video?
    (1 vote)
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  • primosaur ultimate style avatar for user Ashish Ramesh
    I am slightly confused... I understand how he is doing these problems, but just one thing remains unclear: To calculate these volumes, he is essentially finding volume as a function of x (or y) and integrating with respect to x (or y). Wouldn't this just give you the area under the function for volume? How is this equal to the volume itself (i.e. how can we prove that)? Did he mention this in another video?
    (1 vote)
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    • marcimus pink style avatar for user abranagan
      Short answer: because we multiply it to 2pi
      Long answer: if you consider the formula for calculating the area of a cylinder (2pi*r*h) you can think of it that we are finding the area of a rectangle (r*h) and revolving it (2pi) to create the cylindrical shape. The same can be said about the shell method, except we must calculate the area of irregular shapes, which is where our integrals come in! Yes, using these integrals does just find the area between curves but we use this to then revolve it by multiplying it by 2pi.
      (1 vote)
  • leaf green style avatar for user Hiba
    I have a question about volumes of revolution between two graphs. In areas we subtracted the lower from the upper function but I think we can't do this in volume. For example, if we have two functions like f(x)=x and g(x)=x^2-2 the intersection between the two graphs is at x=-1 and x=2 . During this interval f(x)=x lies above g(x)=x^2-2.But when we calculate the integration of pi∫ x^2-(x^2-2)^2 from -1 to 2 we get a negative answer and volume can't be negative ! However if we compare (fx)^2 to (gx)^2 the latter would be the "upper function" and we would get a positive answer . So do we compare fx and gx or (fx)^2 and (gx)^2 before we plug the functions into the rule? Thanks and sorry for the long question !
    (1 vote)
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Video transcript

In the last video, we set up this definite integral to evaluate the volume of the solid of revolution that we set up using the shell method. So now, let's just evaluate this thing. So the main thing is just simplifying this expression. I'll start off by trying to simplify this part of it. So that's going to be y plus 1. I just ate an apple, so something weird just happened in my throat. But anyway, that's done with. y plus 1 minus y squared minus 2y plus 1. I just expanded out this binomial. And then that would simplify to-- another apple in my throat moment-- so that's going to be y plus 1 minus y squared plus 2y minus 1. So this 1 and this negative 1 cancel out. And let's see. You get negative y squared plus 3y. And then we're going to multiply that times y plus 2. So when you multiply y plus 2 times this, so you have y times negative y squared, it gets us negative y to the third power. y times 3y is going to be plus 3y squared. 2 times negative y squared is negative 2y squared. And then 2 times 3y is plus 6y. So then you go all the way down here. This thing can simplify, too, because you have 3y squared minus 2y squared. So this going to be negative y to the third plus-- this part right over here simplifies to just y squared-- y squared plus 6y. So that's this entire part simplified to this down here. We can take the 2 pi out of the integral sign. So let's do that. We're integrating from y is equal to 0 to y is equal to 3 dy. And I took the 2 pi out here, and that is equal to our volume. And so now, we're essentially ready to take the antiderivative. This is going to be equal to 2 pi times the antiderivative of this business evaluated at 3 minus evaluated at 0. And I'll color code it. I found this useful. The antiderivative of y to the third is y to the fourth over 4, so this is negative y to the fourth over 4. Antiderivative of y squared is y to the third over 2-- or y to the third over 3, I should say. And then finally, I'll do it in yellow. Antiderivative of 6y is 3y squared, so plus 3y squared. And we are going to evaluate all of this business at 0 and 3. So this simplifies. This going to be equal to 2 pi times-- well, let's see. Let me do it in the same colors. 3 to the fourth power is 81. So it's negative 81 over 4, plus-- 3 to the third is 27 divided by 3 is 9, plus 9. And then 3 squared is 9 times 3 is 27, plus 27. And then when you evaluate all of these things at 0, you just get 0. So you're just subtracting out 0, so we really don't have to do anything else with the 0. And now we are ready to simplify. Let me see. Actually, let's just add them all up. So this is going to be 9 plus 27 is 36. So that is 36. And if we want to add it to negative 81 over 4, we just have to find a common denominator. So all of this business is going to be equal to 2 pi times-- and so our common denominator can be 4, times something over 4. We have negative 81 over 4, and then 36 times 4 is 144. Is that right? Yeah, that's 144. So 36 times 4, so it's plus 144. 30 times 4 is 120 plus another 24 is 144. So you have 144, essentially, minus 81. So this is going to be equal to 2 pi times-- and actually, I can even simplify it little bit more, because we have a 2 here and a 4 there. So divide the numerator and denominator by 2 so you get it over 2. So you're going to have pi times-- this is going to be 44. Let's see. If this was an 80, this would be 64. So it's going to be 63. Let me write it this way. It's going to be 63 pi over 2. Did I do that right? 60 plus 81 is 141. Add another 3, you get a 144. Yep. And we're done. We figured out the volume of our front of jet engine-looking shape.