For example, when calculating the volume of a solid of revolution for the area between the curves y=-abs(x-4)+4 ; y=0 about the x axis. I know the equation should be set up as pi * integral between 0 and 8 of (-abs(x-4)+4)^2 dx , but am unsure of how to proceed thereafter.
[Math] How to deal with absolute values in a function when calculating volumes of solids of revolution
calculussolid of revolution
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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.
Just to offer a closure to this question, this may benefit from the inclusion of diagrams. Erlend appears to be proposing the use of "disks", but it should be keep in mind that disk "slices" are always perpendicular to the rotation axis. Since that is "vertical" in this problem, the slices are "horizontal" and so will have "thicknesses" $ \ dy \ . $ So we will need to integrate in the $ \ y-$ direction, which requires expressing the disk radii as a function of $ \ y \ . $
The function inversion for the parabola is $ \ x = \sqrt{\frac{y}{a}} \ $ , and this is applicable over the entire interval in the $ \ y-$ direction, $ \ 0 \le y \le h \ , $ so the volume integral is
$$ \int_0^h \ \pi \ [x(y)]^2 \ \ dy \ = \ \pi \ \int_0^h \ \frac{y}{a} \ \ dy \ = \ \frac{\pi}{ a} \ \left( \ \frac{y^2}{2} \vert_0^h \ \right) \ = \ \frac{\pi \ h^2}{2 \ a} \ , $$
as already shown by Jean-Claude Arbaut.
[The apparent discrepancy in "dimensionality" is due to the fact that $ \ a \ $ has the "dimension" of inverse length. Our result is correctly a three-dimensional volume.]
Some of Erlend's confusion may be arising from the requirements for applying the "shell method." For that, the shell-wall "slices" are always parallel to the rotation axis.
The shell "thicknesses" are now $ \ dx \ , $ so the integration is carried out in the $ \ x-$ direction, which will use the single interval $ \ 0 \le x \le \sqrt{\frac{h}{a}} \ , $ as he was writing (but for the wrong method). It is the integrand which must be "split", as the "height" of the shells extends from the parabola "upward" to the "horizontal" line $ \ y = h \ . $ Here, we express the function for the parabola in terms of $ \ x \ , $ making the shell heights $ \ h - ax^2 \ . $
The radii of the shells again extend perpendicularly from the rotation axis, so these are given by $ \ r = x \ . $ The shell volume integration is thus
$$ \int \ 2 \pi \ r \ h \ \ dr \ \ \rightarrow \ \ 2 \pi \ \int_0^{\sqrt{\frac{h}{a}}} \ x \ \cdot \ (h - ax^2) \ \ dx $$
$$ = \ 2 \pi \ \int_0^{\sqrt{\frac{h}{a}}} \ hx - ax^3 \ \ dx \ = \ 2\pi \ \left( \ \frac{hx^2}{2} \ - \ \frac{ay^4}{4} \ \vert_0^{\sqrt{h/a}} \ \right) $$
$$ = \ 2 \pi \ \left( \ \frac{h}{2} \cdot \frac{h}{a} \ - \ \frac{a}{4} \cdot \frac{h^2}{a^2} \ \right) \ = \ 2 \pi \ \cdot \ \frac{h^2}{4a} \ = \ \frac{\pi \ h^2}{2 \ a} \ , $$
as found above.
Best Answer
I've done this volume calculation using Pappus's $(2^{nd})$ Centroid Theorem: the volume of a planar area of revolution is the product of the area $A$ and the length of the path traced by its centroid R, i.e., $2πR$. The volume is simply $V=2\pi RA$.
The figure below shows the area to be revolved about the $x$-axis. The centroid of a triangle is straightforward and is shown here by the asterisk $x_c,y_c=4,4/3$ and the area is simply $A=16$.
Hence,
$$V=2\pi\frac{4}{3}16=\frac{128\pi}{3}$$
I've verified this result numerically by an alternative method. Even if this isn't the method you want, you'll have a (verified) result to compare.