surface area of a solid of revolution. (request for a source for rigourse proof )

Question

link: Doubt in Application of Integration - Calculation of volumes and surface areas of solids of revolution.

I know that this question has been asked and answered before , but non of the answers give a rigourse treatment of the problem . All the explanations boil down to give some argument using infinitesimals and how we should keep the linear orders and neglect the higher orders in the integral. But why should we neglect the non linear terms ? who decides that it's okay to neglect non linear terms, but it's not okay to do the same for the linear terms.What i'm looking for is a rigourse proof for the formula of the surface area of a solid of revolution. Preferable not very advanced, a reference for a book also would work.

Answer 1

I think have a decent outline of a proof. First prove the relationship between area under a curve and definite integration. Then prove how to convert Surface Area and volume problems to related definite integral problems rigorously.

It might be best to start with simple area under the curve of a function. The area under a a curve can be represented by a Riemann Sum. The Sum is easily calculated from usual geometric formulae and we know it will be less than the actual area under the curve.

We can proceed from that in a couple different ways. The simpler geometric figures added up to calculated the area can shrink, requiring more of them. We keep track of the error term and note that it gets arbitrarily small. That should give you a rigorous proof with the dropping of the square terms of the error.

An alternative approach, one can construct a Riemann Sum that deliberately over estimates the area. It can be shown that the overestimate and the underestimate converge using various convergence proofs.

Applying the same principles to 3 dimensions then gives you the usual formulas.

Every definite integral is finding the area of a curve, so whatever 1D integral gives you a volume or a surface area is a problem that has been re-represented as the area under a curve. The function you end up integrating contains the geometric information of the surface of interest. So once you prove that the new function does in fact contain the desired information, then the proofs of integrating a single valued function apply.

A slightly different approach, more geometric approach:

Given a vector function $\vec{E}(x,y,z)$ the volume integral of its Divergence is the surface integral of its component normal to the surface:

Letting $\hat{n}$ be the normal to our surface.

$$\int\int\int \nabla\cdot\vec{E} \ dV = \int\int \vec{E} \cdot \hat{n} \ dA $$

So if the divegrence of $\vec{E}$ is 1, then the integral gives us the entire volume of the region of integration.

Here we note that if $\vec{r}$ is a position vector from the origin to the point of integration, $\nabla \cdot \vec{r}/3=1$. Which means our volume can be expressed as a surface integral $\int\int \vec{r}\cdot \hat{n}/3 \ dA$

Our surface integral is $\int\int 1 \ dA=\int \int \hat{n} \cdot \hat {n} \ dA$. Working in reverse direction from before, we can conver the area integral into a volume integral using the with the reverse of Gauss' Law.

Area = $\int\int \int \nabla \cdot \hat{n} \ dV$

Note the relationship between the Area Integral expression for Volume vs. The area integral expression for surface area. One is proportional to $\hat{n} \cdot \hat{n}$, the other is proportional to $\vec{r} \cdot \hat{n}=r\cos{\theta}$ where $\cos{\theta}$ That cosine term represents a needed transformation to change the area integral to a volume integral, information about the shape of an object being integrated.

Also keep in mind how triple integrals work in Cylindrical Coordinates.

Combine all this I think you get a proof without the "ghost of departed quantities".