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Surface Area and Volume of Solid of Revolution: Why does $\int 2 \times \pi y \, dx$ not work for surface area, but $\int \pi \times y^2 \, dx$ works for volume?

I know that for surface area, it’s because the function is slanted so it couldn’t be written as Riemann sum but also for volume, it can be written as Riemann sum while the function is also slanted.

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  • $\begingroup$ It sounds like you have the basic reason. Do you still have issues understanding the reason? $\endgroup$ – Brian Tung May 11 '17 at 1:41
  • $\begingroup$ I understand the reason. But I'm saying that we can use the same reason to disprove the formula for volume of the solid of revolution (write it with ds instead, like the actual formula for surface area). (I'm not trying to disprove anything but just trying to understand it.) $\endgroup$ – Arina Momajjed May 11 '17 at 1:45
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    $\begingroup$ Consider a cyliner of radius $r$ and height $2r$. It is a solid of revolution, and it has a surface area (not counting the endcaps) of $4 \pi r^2$, and a volume of $2 \pi r^3$. If you were to make the surface bumpy, the volume would barely be affected at all, while the surface area would be increased significantly. $\endgroup$ – Brian Tung May 11 '17 at 7:32
  • $\begingroup$ The reason, in terms of the local "slant" of the surface, is as follows: Suppose you had a local slant of $1$—that is, over a layer of thickness $\Delta x$, the radius went up by $\Delta x$ also. The volume of that layer, which was already $\pi r^2 \Delta x$, goes up by no more than $2 \pi r (\Delta x)^2$; the presence of an infinitesimal, squared, suggests that this increase is small. The surface area of that layer, however, goes up from $2 \pi r \Delta x$ to $2 \pi r \Delta x \sqrt{2}$ (because the cross section of the surface area is the hypotenuse of a 45-90-45 right triangle). $\endgroup$ – Brian Tung May 11 '17 at 7:35
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    $\begingroup$ Possible duplicate of Areas versus volumes of revolution: why does the area require approximation by a cone? $\endgroup$ – Hans Lundmark Nov 25 '19 at 15:16
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Here's an explanation that mixes varying amounts of intuition and slight rigor.


Volume of a solid of revolution:

When using the disk method the idea is that we're adding up the volumes of a massive amount of extremely thin disks between $x=a$ and $x=b$ in order to get the volume of the solid. The disks each have radius given by $y(x)$ and thickness given by $\Delta x$. So the volume of each disk is $\pi [y(x)]^2 \Delta x$. Then we push this through the limit process so that we're using integration to add up the volumes of infinitely many infinitely thin disks. Here, $y(x)$ stays $y(x)$ and $\Delta x$ becomes $dx$ in the integral, and we have $\displaystyle V = \pi \int_a^b [y(x)]^2 \, dx$.

The "slant" of the function is completely irrelevant here.


Surface area of a solid of revolution:

To find the surface area, you want to add up the surface areas of the boundaries of a massive amount of extremely tiny approximate disks. (My use of the word "approximate" will be explained shortly, and until then I'll just keep saying disk and I'll also stop specifying that we only want the surface areas of the boundaries.) Each disk has radius given by $y(x)$. But this time, we can't take $\Delta x$ to be the thickness. Why not? Because $\Delta x$ is a good approximation for the thickness of the interior of the disk, but not for the boundary of the disk. And it's the boundary we care about for surface area, because surface area depends on circumference, and the circumference depends on what's happening on the boundary.

How can we get the thickness (or length) of the boundary? We need to approximate the boundary with tiny line segments. Each line segment can be viewed as the hypotenuse of a tiny right triangle whose legs are parallel to the $x$- and $y$-axes. The leg parallel to the $x$-axis has length $\Delta x$ and the leg parallel to the $y$-axis has length $\Delta y$. If we let $\Delta s$ denote the length of our tiny line segment then we get $\Delta s = \sqrt{(\Delta x)^2 + (\Delta y)^2}.$

Here's a picture illustrating this. Imagine this is some function that we very closely zoomed in on:

enter image description here

So, each of our little tiny disks has surface area approximately given by $2\pi y(x) \sqrt{(\Delta x)^2 + (\Delta y)^2}$. Note that our disks actually have slanted boundaries. (The line segments whose lengths are $\Delta s$ are the boundaries of our disks.) This is why I said "approximate disks" earlier: Our tiny disks for surface area are not perfect disks (like we had for volume) because they're actually disks with slanted boundaries. And this whole slanted boundary thing is necessary here because we must accommodate for the slant of the function, because that slant affects the boundaries of our disks (but not their interiors, which is why we didn't care about slant for the volume).

Anyway, we can rewrite the approximate surface area of each slanted-boundary disk as: $$ 2\pi y(x) \sqrt{(\Delta x)^2 + (\Delta y)^2} = 2\pi y(x) \sqrt{(\Delta x)^2 \left[\frac{(\Delta x)^2}{(\Delta x)^2} + \frac{(\Delta y)^2}{(\Delta x)^2}\right]} = 2\pi y(x) \sqrt{1 + \left(\frac{\Delta y}{\Delta x}\right)^2} \, \Delta x$$

Then we push this through the limit process so that we're using integration to add up the surface areas of infinitely many infinitely thin disks. Here, $y(x)$ stays $y(x)$, and $\Delta y$ becomes $dy$ and $\Delta x$ becomes $dx$. So we get $\displaystyle S = 2\pi \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$.


Summary:

For volume, we start by adding up the volumes of a massive amount of extremely thin disks. Since we're adding volumes, it's the interiors of the disks that we care about. And $\Delta x$ is a good enough approximation for the thickness of each disk if we're looking only at the interiors of the disks.

For surface area, we start by adding up the surface areas of a massive amount of extremely thin disks. Since we're adding surface areas, it's the boundaries of the disks that we care about. But $\Delta x$ is not a good enough approximation for the thickness of each disk, because it doesn't account for the slant of the function. This is why $\Delta s$ is necessary. And the slant of the function is important for this because the function is the boundary of our solid of revolution, and we need to use the thickness of that boundary to get the circumferences of our "disks" so that we can get a working approximation of the surface areas.

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  • $\begingroup$ I like your explanation of why $\Delta s$ is necessary for surface area. But why do you say we care about the interiors of the disks for volumes? And why we can we choose "perfect disks" for volume (whereas we are forced to use "approximate disks" for surface area)? I apologize if I'm asking the original question in different words. $\endgroup$ – SplitInfinity May 11 '17 at 5:58
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You are quite correct in your supposition that there should be some symmetry in the surface area and volume of a body of revolution. Your only error is that in the surface are we do not want the area under $ydx$, but rather along $yds$.

These ideas are expressed quite clearly in Pappus's Centroid Theorems:

Pappus's $(1^{st})$ Centroid Theorem states that the surface area $A$ of a surface of revolution generated by rotating a plane curve $C$ about an axis external to $C$and on the same plane is equal to the product of the arc length $s$ of $C$ and the distance d traveled by its geometric centroid (Pappus's centroid theorem). Simply put, $S=2\pi RL$, where $R$ is the normal distance of the centroid to the axis of revolution and $L$ is curve length. The centroid of a curve is given by

$$\mathbf{R}=\frac{\int \mathbf{r}ds}{\int ds}=\frac{1}{L} \int \mathbf{r}ds$$

Pappus's $(2^{nd})$ Centroid Theorem says 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 bottom line is that the volume is given simply by $V=2πRA$. The centroid of a volume is given by

$$\mathbf{R}=\frac{\int_A \mathbf{r}dA}{\int_A dA}=\frac{1}{A} \int_A \mathbf{r}dA$$

Thus we can say for your cases that

$$ S=2\pi\int y\ ds \\ V=\pi\int y^2\ dx $$

Notice that here, the length, $L$ and the volume, $V$ have cancelled out, thus disguising the true nature of the equations.

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