Models for Determining Volumes and Surface Areas

[whoareyou]

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.

 

[Dick]

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.

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.

 

[whoareyou]

Oh right, I made a mistake in the integral! So then why is this the case?

 

[Dick]

Oh right, I made a mistake in the integral! So then why is this the case?

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.

 

[whoareyou]

Why doesn't this extra space have to be factored into determine the volume?

 

[Dick]

Why doesn't this extra space have to be factored into determine the volume?

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.

 

[whoareyou]

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.

The situation is analogous to the curve-length formula: following the same approach in that case, you would just integrate the width of each infinitesimal segment, giving the very uninteresting integral[dx] = x. The length along the curve of the sloped segment, sqrt(dx^2 + dy^2), does NOT approach dx as dx goes to 0; rather it approaches sqrt(1+(dy/dx)^2)*dx. Therefore this factor must be retained in the curve length formula.

Source: http://mathforum.org/library/drmath/view/51814.html

I'm not understanding this. Doesn't $\displaystyle\lim_{dx\rightarrow 0}\sqrt{dx^2 + dy^2} = \sqrt{dy^2} = dy$ ?

 

[Dick]

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 $\displaystyle\lim_{dx\rightarrow 0}\sqrt{dx^2 + dy^2} = \sqrt{dy^2} = dy$ ?

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.

 

[whoareyou]

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?

 

[Dick]

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?

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?

 

[whoareyou]

If the error is the true area- the approximating area, how would you calculate the true area for some strip of with dx?