Why use different intuitions for volume and surface of revolution.

Question

Suppose $y=f(x)$ is a continuous curve on $[a,b]$.Suppose we are to find the volume of revolution of the solid generated by the area under $f(x)$ and bounded by $x$-axis and the ordinates $x=a$ and $x=b$.

We consider the solid to be composed of elementary cylindrical slices of width $\delta x$ and radius $f(x)$ and we integrate their volumes to get the formula as $\int_a^b \pi [f(x)]^2 dx$.

Note that while we are considering cylindrical slices we are neglecting the curvature of $f(x)$ and the logic is roughly given as $\delta x$ is very small,so the difference between $f(x)$ and $f(x+\delta x)$ is small by continuity and so we can consider it as $f(x)$ only and neglect the curvature.

The logic seems fine.But when we come to calculating the surface of revolution of the surface generated by rotating $f(x)$ curve about $x$-axis,we give some different logic.

We do not consider cylindrical rings of thickness $\delta x$ in this case.Rather we take the curvature of $f(x)$ into account and consider a bend ring as shown in the second diagram below?If we take $\delta x$ thickness and consider cylindrical rings of radius $f(x)$ then I am getting wrong answer.

Why is it so that for volume and area we have different explanations?

Answer 1

When you consider the volume, you can consider for every cylindrical piece the smallest and largest volume. To write this in mathematical notation: $$\pi\ dx\min_{dx}(f(x))^2\le dV\le\pi\ dx\max_{dx}(f(x))^2$$ In the limit $dx\to 0$ the minimum and maximum converge to $dV$.

Similarly, for the area, you need to consider the lateral area of the frustum of height $dx$ and radii $f(x)$ and $f(x+dx)$: $$dA=\pi(f(x)+f(x+dx))\sqrt{(f(x+dx)-fx(x))^2+dx^2}\\=\pi(f(x)+f(x+dx))dx\sqrt{\frac{(f(x+dx)-fx(x))^2}{dx^2}+1}\\\approx2\pi f(x)\sqrt{[f'(x)]^2+1}dx$$

Answer 2

One way to justify this is by parametrizing the surface of revolution. Suppose that $f$ is positive on $[a,b]$ and define $\vec{r}:[a,b]\times[0,2\pi) \rightarrow \mathbb{R}^3$ by $$\vec{r}(u,v)=(u,u\cos(v),u\sin(v))$$ It turns out that $\vec{r}$ is a parametric representation for the surface of revolution with surface area $$\int_0^{2\pi}\int_a^b\|\vec{r}_u\times\vec{r}_v\|dudv=\int_0^{2\pi}\int_a^bf(u)\sqrt{1+\big(f'(u)\big)^2} dudv=\int_a^b2\pi f(u)\sqrt{1+\big(f'(u)\big)^2}du$$ Meanwhile, the solid can be interpreted as the image of $T:[a,b]\times[0,1]\times[0,2\pi)\rightarrow \mathbb{R}^3$ given by $$T(u,v,w)=\big(u,f(u)v\cos(w),f(u)v\sin(w)\big)$$ Its volume necessarily equals the triple integral $$\int_0^{2\pi}\int_0^1\int_a^b\left|\frac{\partial(x,y,z)}{\partial(u,v,w)}\right|dudvdw=\int_0^{2\pi}\int_0^1\int_a^b\big(f(u)\big)^2vdudvdw=\int_a^b\pi\big(f(u)\big)^2du$$