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Models for Determining Volumes and Surface Areas

  1. Jun 21, 2014 #1
    When we generate solid by rotating a curve around an axis, we use "slabs" of cylinders to approximate the volume of this solid of revolution. When we want the find the surface area, we instead use "slabs" of conical frustums (ie. the slope of the differential length of curve is taken into consideration). Why is this?

    The way I see it: When find the area under a curve, we approximate using rectangles. If you were to rotate the curve along with those rectangles, you generate approximating cylinders which can be used to find the volume. So why is it different when trying to find surface area?

    I've tried to find the surface area of a sphere by using cylinders and not the frustums and obtained the correct surface area formula.
     
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  3. Jun 21, 2014 #2

    Dick

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    You will NOT obtain the correct surface area of a sphere using the surface area of enclosed cylinders unless you made a mistake. Post your workings. The cylinders are good approximation to the volume. They don't approximate the surface area well. It doesn't even work for a cone.
     
  4. Jun 21, 2014 #3
    Oh right, I made a mistake in the integral! So then why is this the case?
     
  5. Jun 22, 2014 #4

    Dick

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    Because of the slope of the sides. Take a 45 degree cone and imagine splitting it into equally spaced cylinders. It won't approach the area of the cone no matter how small the cylinders are. You'll be off by a factor of sqrt(2). That's why there is a f'(x) in the formula for area.
     
  6. Jun 23, 2014 #5
    Why doesn't this extra space have to be factored in to determine the volume?
     
  7. Jun 23, 2014 #6

    Dick

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    Because the extra volume approaches 0 as the size of the cylinders decreases. The extra area doesn't. Think of trying to get the arc length of the hypotenuse of a triangle by summing the sides of approximating rectangles. It just doesn't work.
     
  8. Jun 23, 2014 #7
    Intuitively, it makes sense but I can't seem to make it out mathematically and I want to start all the way from the bottom.

    I'm not understanding this. Doesn't [itex]\displaystyle\lim_{dx\rightarrow 0}\sqrt{dx^2 + dy^2} = \sqrt{dy^2} = dy[/itex] ?
     
  9. Jun 23, 2014 #8

    Dick

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    Not if dy depends on dx. If the curve is y(x)=f(x) then dy=f'(x)dx. If f'(x) is nonzero you can't ignore what's happening with dy.
     
  10. Jun 24, 2014 #9
    Ok I see. Now, how would I go about showing that this arc length doesn't affect the volume and we could just use regular cylinders for the integral?
     
  11. Jun 24, 2014 #10

    Dick

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    Try to think of an argument why. You have the approximating cylinder with volume dV=pi*f(x)^2*dx. Can you think of an upper bound for the amount error (call it dE) you are making in neglecting the true shape of the curve? Can you show dE/dV goes to zero as dx approaches 0?
     
  12. Jun 27, 2014 #11
    If the error is the true area- the approximating area, how would you calculate the true area for some strip of with dx?
     
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