# 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?

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