# Why use different intuitions for volume and surface of revolution.

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?

## 2 Answers

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 $$f(x+dx)$$: $$dA=\pi(f(x)+f(x+dx))\sqrt{(f(x+dx)-fx(x))^2+dx^2}\\=\pi(f(x)+f(x+dx))dx\sqrt{\frac{(f(x+dx)-fx(x))^2}{dx^2}+1}\\\approx2\pi f(x)\sqrt{[f'(x)]^2+1}dx$$

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 $$\int_0^{2\pi}\int_0^1\int_a^b\left|\frac{\partial(x,y,z)}{\partial(u,v,w)}\right|dudvdw=\int_0^{2\pi}\int_0^1\int_a^b\big(f(u)\big)^2vdudvdw=\int_a^b\pi\big(f(u)\big)^2du$$