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

Lesson 11: Shell method

# Shell method with two functions of y

Stepping it up a notch, our solid is now defined in terms of two separate functions. Created by Sal Khan.

## Want to join the conversation?

• ishh why is the lenghth? of the the slice bottom function minus top function?? (y+1-(y-1)^2. Everything up to this point makes sense but now Im lost. . . I thought you do top minus bottom/ right most - left most
• Tilt your head 90 degrees to the right. Which is upper function now?
If we calculate in terms of y then the function output is x, and x=y+1 returns higher x'es then x=(y-1)^2, thus x=y+1 is the top function.
• at , how is it y+2 for the radius?
• In other words, it is because the shape is not rotated around the x axis but rather around a line that is two units below the x-axis. Therefore to get the radius, you have to add those two units to the value of y, so, (y + 2) is the radius
• After the mark, what does he do (y+1 - (y-1)^2) instead of the other way around? is x=(y-1)^2 not the larger radius, and he should therefore subract the smaller y+1 function to get the area between the two?
• We are integrating with respect to y, so the functions are in terms of y. To visualize this, tilt your head to the right (as Sal said) and look at the graph. Then you'll see y+1 is on top while (y-1)^2 is on the bottom. So in this case, y+1 is the larger radius.
• Is the radius always positive? Like, what if the axis of rotation was y=2 instead of y=-2? Would the radius still be y+2?
• A radius is a physical length, so it can't really have a negative value. No matter what the axis of rotation is, the radius will always be positive. :)
(1 vote)
• So disk method is using area and shell method is using circumference ?
• , he denotes the distance from the x axis to the yellow function as y, but why does he use the upper function? Wouldn't there technically be two y's, one to the yellow and one to the blue function (and I realize that the yellow curve isn't an actual function, but I'm just calling it that for the sake of simplicity).
• Good question! It does look like he is defining y as the distance from the x-axis to the yellow function because of where he draws the vertical purple segment, but really it is giving the distance to the entire thin horizontal pink rectangle, not just to the point on the yellow curve. He could (and for clarity, he probably should) have drawn the purple segments a bit further to the right, to the middle of the thin rectangle.

Because the thin slices run horizontal, the y-values of the two curves are not what get subtracted, but instead we need to subtract the x-values of the curves. That's what is happening around , when he talks about the "upper function" and the "lower function". Like you said, the yellow curve is NOT a function when we think of y as a function of x, but it IS a function if we consider its x-value as a function of y. I don't think he should refer to them as "upper" and "lower". Instead, I think he should call the blue curve the "right" curve and the yellow curve the "left" curve. So, to get the length of the thin rectangular slices, he does the x-value of the RIGHT curve (x = y + 1) minus the x-value of the LEFT curve (x = (y-1)²).
(1 vote)
• At - Out of curiosity, if I swapped the upper and lower functions, would it just result in a negative volume?
• I'm guessing it would be like putting an upper bound on the bottom of the integral which results in putting a negative sign infront of the integral to make the upper bound go on top; so I think basically yes
(1 vote)
• Why is the thickness dy?
Shouldn't the thickness of shell be dx?
• dy is infinitely small

this is the shell method for functions of y, so we use dy here
(1 vote)
• How could I solve this using disk method? Thanks:)
Here is what I written ,but I typed that into a calculator the answer wasn't correct

sum(0,4) pi (sqrt(x)+3)^2 dx
minus
sum(0,4) pi (x+1)^2 dx
minus
sum(0,1) pi 2^2 dx
plus
sum(0,1) pi (x+1)^2 dx
(1 vote)
• I'm not entirely sure how you built your integrals, but it looks like there might be one major problem keeping you from the right answer. There's a change at x=1 where the lower function changes, so you'll have to adjust your integrands and limits of integration to reflect that. This should work:

For x on the interval (0,1), the upper function is `y=sqrt(x)+1` and the lower function is `y=-sqrt(x)+1`, so the volume for this section (rotated about y=-2) will be given by `int(0,1) [ pi*(sqrt(x)+3)^2 - pi*(-sqrt(x)+3)^2 ] dx`.

For x on the interval (1,4), the upper function is still `y=sqrt(x)+1`, but the lower function is now `y=x-1`, so the volume for this section will be given by `int(1,4) [ pi*(sqrt(x)+3)^2 - pi*(x+1)^2 ] dx`.

The total volume will be the sum of those two integrals. Hope that helps.