I am trying to derive the formula for the area of a sphere using integration. It is coming out as $ \pi^2R^2 $ instead of $4\pi R^2$.

This is what I am doing :-

I am approximating the area of the sphere of radius R (kept at origin) using the Curved Surface Area of infinite infinitesimal cylinders along the X axis.

Now each infinitesimally small cylinder's Curved Surface Area is $ 2π f(x) dx $.

Therefore, Area of the sphere is :-

$$ \int_{-R}^R 2\pi f(x) dx \\ = 2\pi\int_{-R}^R \sqrt{R^2 - x^2} dx \\ \text{Put } x = R \sin \theta \text{, we get} \\ \text{Area} = 2\pi \int_{-\frac{\pi}{2}}^\frac{\pi}{2} R \cos \theta. R \cos \theta d\theta \\ = 2 \pi R^2 \int_{-\frac{\pi}{2}}^\frac{\pi}{2} \cos^2 \theta d \theta \\ = 2 \pi R^2 \int_{-\frac{\pi}{2}}^\frac{\pi}{2} \frac{1 + \cos2\theta}{2} d\theta \\ = \pi R^2 \left [\theta + \frac{\sin2\theta}{2} \right ]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \\ = \pi^2 R^2 $$

What am I doing wrong?

  • $\begingroup$ why you've put $f(x) = \sqrt{R^2 - x^2}$ ? $\endgroup$ – Ajay Mishra Jul 1 '19 at 17:52
  • 5
    $\begingroup$ You need to take into account that the slices are not really cylinders, since their sides are not vertical. Near the top of the sphere, they are very far from vertical, so treating them as vertical will significantly underestimate their surface area. Try using the surface area of a frustum instead. $\endgroup$ – Nate Eldredge Jul 1 '19 at 17:54
  • 1
    $\begingroup$ The same reasoning lets you get that the length of the line segment $(0,0)$ to $(1,1)$ is $1,$ since you can approximate each segment fo4 $(x,x)$ to $(x+\Delta x,x+\Delta x)$ with the length $\Delta x$. $\endgroup$ – Thomas Andrews Jul 1 '19 at 17:59
  • $\begingroup$ But why doesn't the same problem happen when estimating the area under the line segment from (0,0) to (1,1) (and above the x-axis) using rectangles? The slope is quite far from horizontal everywhere along the line segment. $\endgroup$ – Ted Jul 1 '19 at 18:01
  • $\begingroup$ Because we can prove that the error there as $\Delta x\to 0$ goes to zero. @Ted21 $\endgroup$ – Thomas Andrews Jul 1 '19 at 18:02

On the interval $(x_1,x_2)$ the better approximation is not the cone, which misses a lot of area of $f'(x)$ is large on the interval.

Rather, if we think of the section as approximated by a cone, then we have (with $\Delta x=x_2-x_1,$ and $\Delta f = f(x_2)-f(x_1))$ that the point of the cone must be $(c(x),0)$ where $$c(x)=x-f(x)\frac{\Delta x}{\Delta f}$$

Then the area of this region of the one is:

$$\left|\pi\cdot f(x_2)\sqrt{\left(\Delta x+f(x_1)\frac{\Delta x}{\Delta f}\right)^2+f(x_2)^2}-\pi\cdot f(x_1)\sqrt{\left(f(x_1)\frac{\Delta x}{\Delta f}\right)^2+f(x_1)^2}\right|$$

This can be written is:

$$\pi\frac{\Delta x}{\Delta f}\left| \left(f(x_2)^2- f(x_1)^2\right)\sqrt{1+\left(\frac{\Delta f}{\Delta x}\right)^2}\right|$$

Assuming $f'(x_1)\neq 0$ then as $x_2\to x_1$ you get:

$$2\pi \Delta x\left|f(x_1)\sqrt{1+f'(x_1)^2}\right|$$

But this is the value gives the cylinder when $f'(x_1)=0.$

So you get the integral:

$$2\pi \int_{-R}^R f(x)\sqrt{1+f'(x)^2}\,dx$$

Note that with $f(x)=\sqrt{R^2-x^2},$ you have $f'(x)=\frac{-x}{f(x)}$ and $f(x)$ is positive, you can rewrite this as:

$$2\pi \int_{-R}^R \sqrt{f(x)^2+x^2}\,dx=2R\pi\int_{-R}^{R}\,dx=4R^2\pi$$

  • $\begingroup$ I still don't understand why the Riemann sum for surface area of cylinders doesn't work, but for the volume it works, when doing the same thing. $\endgroup$ – arandomguy Jul 2 '19 at 11:18
  • $\begingroup$ Consider the cone gotten from rotating the curved $y=mx$ for $x\in[0,1].$ Geometry tells us that the surface are is $\pi r\sqrt{r^2+h^2}=\pi m\sqrt{m^2+1}.$ your approximation with cyllnders would be $\int_0^1 2\pi mx\, dx =\pi m.$ $\endgroup$ – Thomas Andrews Jul 3 '19 at 23:56
  • $\begingroup$ @arandomguy When the slope of a function is very large at $x,$ the rotation surface area is closer to an annulus of radii $f(x)$ and $f(x)+f’(x)dx $ which has area $2\pi f(x)f’(x)dx,$ not the cylinder estimate $2\pi f(x)dx.$ In the case of the sphere, you contribute hardly any are for $x$ near $-R,$ but that are contributes considerable are due to the slope (which, in this case, goes to infinity.) $\endgroup$ – Thomas Andrews Jul 4 '19 at 0:07

Since $x^2+y^2=R^2$,$$xdx+ydy=0\implies\frac{dy}{dx}=-\frac{x}{y}\implies\frac{ds}{dx}=\sqrt{1+\left(\frac{dy}{dx}\right)^2}=\frac{R}{y}.$$The surface is the integral of $2\pi yds$, not $2\pi y dx$. For an intuitive understanding of why, see this video's comparison of the sphere's surface to a cylinder's. (It'll even help you compare them without ordinary calculus.) Since $yds=Rdx$, the final result is$$2\pi R\int_{x=-R}^{x=R}dx=4\pi R^2.$$


As pointed out by Nate in the comments, your approach for deriving the problem need a slight modification, as the sphere have a curvature in vertical direction unlike the cylinders. Take your infinitesimal surface as frustum, with slanted length $R d \theta$ (?) and circumference $ 2 \pi R \sin \theta$ (?) $$\implies dA = 2 \pi R^2 \sin \theta d\theta \implies A =2 \pi \int^{\pi}_0 R^2 \sin \theta d \theta$$

  • $\begingroup$ Why does that matter? Since dx approaches 0, then the cylinders should become exact circles. Then why wouldn't I get exact circumference of the circle at every x ( in other words the area of the sphere) ? $\endgroup$ – arandomguy Jul 1 '19 at 19:27
  • 3
    $\begingroup$ Briefly: When you use approximations, you must estimate the error. If you use your approximation (cutting into $N$ cylinders of width $\delta x$), the error in the surface area in each slice is on the order of $\delta x$. Thus the total error is $N \cdot \delta x \approx 1$, so the error does not go to 0 as $\delta x \to 0$, so the integral does not represent the surface area. To see the error more clearly, work out the case of a line segment as in Thomas Andrews' comment. $\endgroup$ – Ted Jul 1 '19 at 19:55
  • $\begingroup$ Here's a variant of what you're struggling with. $\endgroup$ – Raskolnikov Jul 1 '19 at 20:25
  • $\begingroup$ @Ted Why doesn't the error go to 0 if dx -> 0 ? We are using integration here $\endgroup$ – arandomguy Jul 2 '19 at 12:19
  • $\begingroup$ Because you are cutting up a fixed size interval (say of length 1) into $N$ pieces each of length $\delta x$. If $\delta x \to 0$, then at the same time, $N \to \infty$. So $\delta x \cdot N$ stays at a constant 1, even as $\delta x \to 0$. You need the total error to go to 0, not just the error for an individual slice. $\endgroup$ – Ted Jul 2 '19 at 17:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.