Same question as this but for surface area instead of volume: Volume of revolution on an area crossing the axis

1. Is this correct to compute the volume of the solid of revolution?

To compute the volume of the solid of revolution obtained by revolving the area of the region (just the regular use of the word 'region' and not region in topology) between functions $f$ and $g$ and between $x=a$ and $x=b$ around $x$-axis, where $a<b$ and where the region crosses the $x$-axis, we use the function $h := \max\{|f|,|g|\}$

$$V = \pi \int_{a}^{b} (h(x))^2 dx = \pi \int_{a}^{b} (\max\{|f(x)|,|g(x)|\})^2 dx$$

$$= \pi \sum_{i=0}^{n-1}\int_{p_i}^{p_{i+1}} (\max\{|f(x)|,|g(x)|\})^2 dx$$

over some partition $\{p_0,...p_{n}\}$ of $[a,b]$ where $h(x)$ changes from one element of the partition to the next.

2. To compute the surface area of the solid of revolution, what $h$ do we use?

To compute the surface area of the solid of revolution obtained by revolving the area of the region (just the regular use of the word 'region' and not region in topology) between functions $f$ and $g$ and between $x=a$ and $x=b$ around $x$-axis, where $a<b$ and where the region crosses the $x$-axis, what $h$ do we use?

$$SA = 2 \pi \int_{a}^{b} h(x)\sqrt{1+(h'(x))^2} dx = 2 \pi \int_{a}^{b} ? \sqrt{1+(\frac{d}{dx}?)^2} dx $$

  • Is it still $\max$? Of course, I'm assuming overlap is still a problem in surface areas of solids of revolution just as it is a problem in computing volumes of solids of of revolution.

  • If there's an answer out there, then you don't have to justify the answer: please just link to where I can find the answer, and I'll understand it on my own.

3. Where can I find examples or even definitions of these?

I actually couldn't find any examples in Calculus by James Stewart for either question.


Clearly, we can always translate our functions so as to reduce to the case where the rotation is done around the $x$ axis, with $a \le x \le b$.

We assume then that we are speaking of rotation of $2 \pi$.
The case of a rotation of only $\pi$ radians is a bit different.
We also concentrate the discussion on the lateral surface, leaving out the surface at the basis $a$ and $b$ whose computation is easy.


Consider the case 1) in which $$ 0 \le g(x) \le f(x)\quad \left| {\,a \le x \le b} \right. $$ When rotating of $2\pi$, the "external" curve will generate a volume which is considered positive same as the corresponding lateral surface. The "internal" curve $g(x)$ is instead generating a cavity whose volume is taken as negative (to subtract from the outer one), while the lateral surface is taken to be positive in the geometrical sense.
That is the same as saying that the solid generated is the toroid having a cross-section corresponding to the area comprised between the two curves.

Consider now the case 2) in which $$ g(x) < 0\;\; \wedge \,\;0 \le \left| {g(x)} \right| \le f(x)\quad \left| {\,a \le x \le b} \right. $$ that is, when the rotation axis crosses the area between the two curves.

There is not an unanimous interpretation of what is the rotation solid in this case.
Three commonly occurring interpretations are:

a) "Geometric": the solid is the set of the points swept by the cross-area.
In this interpretation we are considering only the area swept by the larger between $|f(x)|$ and $|g(x)|$, $|f(x)|$ in our example. The solid ( volume and relative surface) generated by $|g(x)|$ is absorbed into that of the larger curve and is therefore not considered.
It is the volume that the rotating section would carve into a soft solid block.

b) "Algebraic": the solid is generated by $|f(x)|, \, |g(x)|$ same as in case 1).
That is the same as to consider negative the portion of the cross-section below the axis (being $|g(x)| \le |f(x)|$). It is also equivalent to apply the Pappus' theorem also when the axis traverses the cross-area. If the theorem is kept also for the surface, the lower portion of the cross-area will provide a negative contribution for that as well.

c) "Physical": both sides of the cross-section contribute positively.
It is the case, for instance, of when the section rotates in a fluid and we are speaking of the volume of the fluid swept.

Under the "Geometric" interpretation, the approach in the general case then follows to be:

  • split the concerned x interval into the intervals for which $$ \left[ {x:\;0 \le \left| {g(x)} \right| \le \left| {f(x)} \right|} \right]\;\; \cup \left[ {x:\;0 \le \left| {f(x)} \right| < \left| {g(x)} \right|} \right] $$

  • for the first type $$ \eqalign{ & 0 \le \left| {g(x)} \right| \le \left| {f(x)} \right|\quad \Rightarrow \cr & \Rightarrow \left\{ \matrix{ {\rm sign}\left( {g(x)} \right) = {\rm sign}\left( {f(x)} \right)\; \Rightarrow \left\{ \matrix{ V = V_{\,f} - V_{\,g} \hfill \cr S = S_{\,f} + S_{\,g} \hfill \cr} \right. \hfill \cr {\rm sign}\left( {g(x)} \right) \ne {\rm sign}\left( {f(x)} \right)\; \Rightarrow \left\{ \matrix{ V = V_{\,f} \hfill \cr S = S_{\,f} \hfill \cr} \right. \hfill \cr} \right. \cr} $$

  • for the second type, same as above exchanging $f(x)$ with $g(x)

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    $\begingroup$ @Mitjackson: a) will work out an example b) sorry, it's a long time that I left dealing with these problems: don't have ref. ready at hand c) in a complete revolution of $2\pi$ , in case 2), it is "normally" understood that the volume/surface swept by both $f(x)$ and $g(x)$ , so two times, is taken as generated once for all. $\endgroup$ – G Cab Mar 2 at 13:18
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    $\begingroup$ @Mitjackson: I added a better explanation. Wish you are convinced about, because at the moment I do not have "academic" references to offer. $\endgroup$ – G Cab Mar 2 at 15:32
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    $\begingroup$ @Mitjackson: by "not accounted" I mean "not taken into consideration", so same as "cancelled", i.e. same if we had $y=0$ in fact. $\endgroup$ – G Cab Mar 2 at 18:52
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    $\begingroup$ @Mitjackson: I re-casted my answer, to make the exposition more organic. $\endgroup$ – G Cab Mar 2 at 23:01
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    $\begingroup$ @Mitjackson: thanks for the bounty: wish the explanation was convincing even without references. $\endgroup$ – G Cab Mar 8 at 12:02

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