# Doubt in Application of Integration - Calculation of volumes and surface areas of solids of revolution

I am using two books for my calculus refresher.

1. Thomas' Calculus
2. Higher Math for Beginners by Ya. B. Zeldovich

My question is : When applying Integral Calculus for calculation of volumes of solids, generated by curves revolved around an axis, we use slices of 'cylinders' to approximate the volume of the resulting solid and then integrate the sum of those infinitesimal cylinders. However, when we are using the same techniques to calculate the surface area of the surfaces generated by revolving curves around an axis, we consider the 'slope' of the differential length of curve 'dx', calculate the total length of the curve and derive the required surface area.

Are we not supposed to use the same 'slope' for calculating the volumes of the infinitesimal 'cylinders' for calculation of volumes? Shouldn't we use 'sliced portions of 'cones' as the infinitesimal volumes?? When it come to calculation of volumes of solids of revolution, why are we neglecting the slope of the curve for the differential length and simply assume that it is an infinitesimal cylinder??

Ex: Let us say we want to calculate the surface area and the volume of the solid generated when the parabola y = 10 . x^2 is revolved about the y-axis, with limits of x from 0 to 5.

In such cases, when calculating the volume of the solid, we consider infinitesimal 'cylinders', ignoring the 'slope' of the curve for the differential element 'dx', but when calculating the surface area, we consider the 'slope' of the differential element 'dx'.

• It may also help to consider the analogous problem in 2D: area under a curve vs. length of the curve over $[a,b]$. In the area case, you can go with either trapezoid or rectangle approximations to set up the integral, but the rectangle approximation is simpler and so we normally use it. However, in order to find curve length, you have to use little triangle hypotenuses (tops of trapezoids) and not horizontal line segments (tops of rectangles). In the latter case you would end up with $b-a$ rather than the actual curve length. – Mike Spivey Sep 7 '11 at 16:04
• Related (but sort of the opposite): math.stackexchange.com/questions/6979/… – Hans Lundmark Nov 25 at 15:24

I like this question; what you're trying to understand is important to understand.

In this answer I'll be talking loosely about infinitesimal quantities linear or quadratic in $\mathrm dx$; I think this is the best way to get a feel for this sort of thing, but similar arguments could also be presented more rigorously.

Basically, the reason is that in the case of the surface area, the effect from the slope is linear in $\mathrm dx$, whereas in the case of the volume, it's quadratic in $\mathrm dx$. Thus we can neglect it in the limit $\mathrm dx\to0$ in the latter case but not in the former.

Let's see what happens if we take the slope into account in adding up the volume of slices of the solid of revolution generated by a function $f(x)$ rotated around the $x$ axis. As you say, after the cylindrical volume $\pi f^2\mathrm dx$ the next order of approximation would be a cone, or more precisely a conical frustum, corresponding to a linear approximation to the function. The volume of such a frustum between $x$ and $x+\mathrm dx$ would be

$$\begin{eqnarray} \frac13\pi\mathrm dx\left(f(x)^2+f(x)f(x+\mathrm dx)+f(x+\mathrm dx)^2\right) &\approx& \frac13\pi\mathrm dx\left(3f(x)^2+3f(x)\mathrm dx\right) \\ &=& \pi\mathrm dx\left(f(x)^2+f(x)\mathrm dx\right)\;, \end{eqnarray}$$

which differs from the cylindrical volume by the second term, which contains one more factor of $\mathrm dx$ than the first one and therefore vanishes in the limit.

By contrast, for the surface area, taking into account the slope leads to a surface element $2\pi f(x)\sqrt{1+f'(x)^2}$, whereas not taking it into account would lead to just $2\pi f(x)$, the surface area of a cylindrical slice. Here we don't have two terms with one negligible and dominated by the other, but an additional factor that survives the limit.

You can also try to picture this geometrically. Think of a conical slice and the corresponding cylindrical slice, and imagine shrinking their width. As you shrink, the portion of volume in that little extra bit on the boundary becomes negligible compared to the bulk of the slice -- whereas the bulk only shrinks with the width, the extra bit shrinks both with the width and with the vertical deviation, which is the slope times the width, so it shrinks quadratically while the bulk shrinks linearly. For the surface, there's no such effect, since there's no "bulk" of the surface; all of the surface is at the boundary, and tilting it by the slope makes all of it larger, not just a small portion that becomes negligible in the limit.

You could use "sliced portions of cones" as your infinitesimal volumes, but the answer would the same as if you used cylinders -- the difference between the two tends to zero faster than the volume itself, so it disappears in the limit. This is not the case with the surface area of a sliced portion of a cone -- its area is greater than the area of a cylindrical slice by a factor that tends to $\sqrt{1+(dy/dx)^2}$ in the limit.