Surface area of a sphere with integration of disks

Question

Why it is not correct to say that the surface area of a sphere is:

$$ 2 \int_{0}^{R} 2\pi r \text{ } dr $$

In my mind we are summing up the perimeters of disks from $r=0$ to $r=R$, so by 1 integration, we would have $\frac{1}{2}$ of the surface area of the sphere.

I know that it's not correct because that will give us $2\pi R^2$ that it's different from $4\pi R^2$, but why???

Thanks!

Answer 1

In order to give a short explanation of the mistake :

In writing $\quad 2 \int_{0}^{R} 2\pi r \text{ } dr \quad$ you sum elementary areas $\quad dA=2\pi\:r\:dr\quad$

that is to say, a strip length$=2\pi r$ and width$=dr$.

The hitch is that the width is not $dr$ but is $ds$ (see the figure).

With a few calculus, you can find $\quad ds=\frac{R}{\sqrt{R^2-r^2}}dr \quad\to\quad dA=2\pi\:r\frac{R}{\sqrt{R^2-r^2}}dr $

$$\quad 2 \int_{0}^{R} 2\pi\:r\frac{R}{\sqrt{R^2-r^2}}dr =4\pi R^2$$

This is a method to find the area of the sphere. Of course, a simpler method consists in doing the job in spherical coordinates instead of Cartesian coordinates.

Answer 2

What you are doing is indeed integrating over rings, but then you get the area of a full circle, which was the result of your calculation up to the irrelevant factor of 2.

You can imagine that by adding more and more dense rings you fill a circle with radius $R$ (the upper limit of your integral).

But to get the area of a sphere, you need to parametrize all 3 dimensions. What you can do is use spherical coordinates, where each point in space is described by a radius $r$, and two angles $\theta,\phi$ as in the link.

In fact, to get the area of the sphere you need to keep the radius constant, and integrate over the angles that parametrize the given sphere.

The result yields

$$S=\int_0^{\pi}\int_0^{2\pi}R^2\sin\theta d\phi d\theta=R^22\pi\int_0^{\pi}\sin\theta d\theta=4\pi R^2$$ as it should be.

Answer 3

$$\text{Area will be}~~: 2 \int_{0}^{R} 2\pi x ~ ds$$

Where $ds$ is width of strip bounded by circles of radius $x$ and $x+dx$ situated at height $y$. Also $ds \neq dr$ it's tilted in $y$ direction too. Only horizontal projection of $ds$ is $dr$.What you have done is valid for Disk See image below.

$$(ds)^2=(dx)^2+(dy)^2$$

Therefore : $$ds = \sqrt{1+\Bigg( \frac{dy}{dx}\Bigg)^2} \; dx$$ Now,

$$x^2 + y^2 = R^2$$ $$y = \sqrt{R^2 - x^2}$$ $$\dfrac{dy}{dx} = \dfrac{-2x}{2\sqrt{R^2 - x^2}}$$ $$\dfrac{dy}{dx} = \dfrac{-x}{\sqrt{R^2 - x^2}}$$ $$\left( \dfrac{dy}{dx} \right)^2 = \dfrac{x^2}{R^2 - x^2}$$

Thus, $$\displaystyle A = 4\pi \int_0^R x \sqrt{1 + \dfrac{x^2}{R^2 - x^2}} \, dx$$ $$\displaystyle A = 4\pi \int_0^R x \sqrt{\dfrac{(R^2 - x^2) + x^2}{R^2 - x^2}} \, dx$$ $$\displaystyle A = 4\pi \int_0^R x \sqrt{\dfrac{R^2}{R^2 - x^2}} \, dx$$

Let $$x = R \sin θ \implies dx = R \cos θ dθ$$

When $x = 0, θ = 0$ When $x = R, θ = \pi/2$

Thus, $$\displaystyle A = 4\pi \int_0^{\pi/2} R \sin \theta \sqrt{\dfrac{R^2}{R^2 - R^2 \sin^2 \theta}} \, (R \cos \theta \, d\theta)$$ $$\displaystyle A = 4\pi \int_0^{\pi/2} R^2 \sin \theta \cos \theta\sqrt{\dfrac{R^2}{R^2(1 - \sin^2 \theta)}} \, d\theta$$ $$\displaystyle A = 4\pi R^2 \int_0^{\pi/2} \sin \theta \cos \theta\sqrt{\dfrac{1}{\cos^2 \theta}} \, d\theta$$ $$\displaystyle A = 4\pi R^2 \int_0^{\pi/2} \sin \theta \cos \theta \left( \dfrac{1}{\cos \theta} \right) \, d\theta$$ $$\displaystyle A = 4\pi R^2 \int_0^{\pi/2} \sin \theta \, d\theta$$ $$A = 4\pi R^2 \bigg[-\cos \theta \bigg]_0^{\pi/2}$$ $$A = 4\pi R^2 \bigg[-\cos \frac{1}{2}\pi + \cos 0 \bigg]$$ $$A = 4\pi R^2 \bigg[ -0 + 1 \bigg]$$ $$A = 4\pi R^2$$

Answer 4

A) when you pile up the disks you shall pay attention to law with which $r$ varies (with $h$, $x$, $\alpha$ $\cdots$) otherwise, as already told you do not make any distinction with a cone, a cylinder,..
B) adding disks to find volumes is ok, because the volume of the triangular cross-section ring that is left-out is a higher order infinitesimal vs. the volume of the disk; its surface instead is of the same order, and you must consider the hypothenuse instead of the cathetus.

Answer 5

If you consider a formula like

$$ dA = 2 \pi r \, dr$$

for a cylinder, it is zero and so fails.

If you consider a formula for a cylinder,

$$ dA = 2 \pi r \, dz$$

it seems to be correct.

However, for a cone $dz=0$ so even this formula fails.

Intuition says $dr,dz$ should both be involved.

In fact if you recall cone formula $ A = \pi\,r\, s = 2 \pi\, \bar {r} \, s ,\, dA= 2\pi r ds $

slant generator is correct in principle.

We have cone differential area

$$ dA = 2 \pi r ds = 2 \pi r \sqrt{dr^2+dz^2 } $$

and that should be considered in integration of all surfaces of revolution.

Answer 6

There are two problems here.

First: Not only do your disks change radius from the apex to the base of the hemisphere, but they do so at a varying rate. Let's say (for concreteness) that the sphere is $x^2 + y^2 + z^2 = R^2$, and we're dividing the sphere into disks based on the $x$-coordinate. If you try to approximate the surface area of a hemisphere of radius $R=1$ by slicing it into disks of radius (say) $0.1, 0.2, \ldots, 0.9, 1$, you'll find that the first few small-radius slices are very close together— to be precise, at $x = \sqrt{1 - 0.1^2} = 0.994$, $\sqrt{1-0.2^2} = 0.980$, $\sqrt{1-0.3^2} = 0.954$, and so forth—but the larger slices are much farther apart. If you use the surface areas of these disks to calculate the surface area of the sphere, you have to take into account the fact that the disks have different widths.

Going from finite to infinitesimal width: The radius of a disk at $x$ is $r = \sqrt{R^2-x^2}$ (the inverse of this formula is $x = \sqrt{R^2 - r^2}$), so if you make cuts in the sphere each time the radius changes by $dr$, then the slices will have width $-dx = -dr \dfrac{dx}{dr} = \dfrac{r}{\sqrt{R^2-r^2}}\, dr$ (the negative sign corrects for the fact that $x$ increases with decreasing $r$).

Second: The surfaces of the sphere slices that we discuss above have sloping cross-sections (in the $xy$ plane or any plane that includes the $x$-axis), whereas cylinders have flat surface cross-sections. You can ignore the difference this makes for calculating volume, but not for surface area; otherwise you can prove that $\pi = 4$. So you need to make another correction: the width of the surface of a disk isn't $dx$ but rather $\sqrt{(dx)^2 + (dr)^2} = \sqrt{ \dfrac{r^2}{R^2 - r^2} + 1}\, dr = \dfrac{R}{\sqrt{R^2 - r^2}}\, dr$. Putting this all together and integrating over $r$ (though $x$ would be nicer): The integral that you actually have to do is thus $$\begin{align*} &\int_0^R (2\pi r) \frac{R}{\sqrt{R^2 - r^2}}\, dr \\ =& \;2\pi R \int_{r=0}^{r=R} \frac{-du/2}{\sqrt{u}} \tag{$u = R^2 - r^2$; $du = -2r\, dr$} \\ =& -2\pi R \left. \sqrt{u} \right|_{r=0}^{r=R} \\ =& -2\pi R \left. \sqrt{R^2 - r^2}\right|_{r=0}^{r=R} \\ =&\;2\pi R^2 \end{align*}$$ and this is the result you want.