# An Integral Error

I was studying the derivations for the volume and surface area of a sphere .

One derivation , for the volume of the sphere is the disk-method .

A circle of radius $$r$$ is considered , centred at origin . The equation of the circle is $$x^2+y^2=r^2$$ .

This implies that $$y^2=r^2-x^2$$ .

To find the volume of the sphere , $$\int _{-r} ^r \pi y^2 \cdot dy$$ = $$\int _{-r} ^r \pi (r^2-x^2) \cdot dx$$ is calculated . This yields $$\frac{4}{3} \pi r^3$$ , the correct answer .

I wondered if a similar method could be used to calculate the surface area of the sphere . So I calculated $$\int ^r _{-r} 2 \pi y \cdot dy$$ = $$\int ^r _{-r} 2 \pi \sqrt{r^2-x^2} \cdot dx$$ . Theoretically , to me , this meant summing up the circumferences , which must yield the surface area .

However , the definite integral evaluates to $$\pi^2 r^2$$ , which is not the correct answer. I would like to know how my approach is wrong , and if at all the area can be calculated by summing up the circumferences (without the use of $$\theta$$ please ) .

Note:- I am aware of other proofs for the surface area , and am not looking for them :) .

• The integral for the area of a circle would be this: $$2\int_{-r}^{r}\sqrt{r^2-x^2}\,dx$$ and for the volume using the disk method, it would be this: $$\pi\int_{-r}^{r}(r^2-x^2)\,dx$$ – Michael Rybkin Feb 18 '19 at 17:01
• @MikeR. Edited . I was looking for a way to find the surface area of the sphere using the disk/ring method , as I have mentioned in the question . – Aspirant Feb 18 '19 at 17:03
• – Aretino Feb 18 '19 at 22:55

The way you're doing it is not the way it's done. In order to find the surface area of a circle, you need to explore the topic of surfaces of revolution. There, you will learn that the formula for surfaces of revolution for two-dimensional objects (for things in the Cartesian plane, in other words) looks like this:

$$S=2\pi\int_{a}^{b}f(x)\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx$$

In that formula, $$f(x)$$ is your curve and $$\frac{dy}{dx}$$ is the first derivative of that curve. In your case, the curve is going to be given by the function $$f(x)=\sqrt{r^2-x^2}$$ whose derivative is:

$$\frac{dy}{dx}=\frac{d}{dx}\left(\sqrt{r^2-x^2}\right)=-\frac{x}{\sqrt{r^2-x^2}}$$

And then you just plug all that information into the formula and integrate (notice the result of the integration should be multiplied by $$2$$ since integration from $$0$$ to $$r$$ will only give us the surface area of a hemisphere):

$$S=2\cdot2\pi\int_{0}^{r}\sqrt{r^2-x^2}\sqrt{1+\left(-\frac{x}{\sqrt{r^2-x^2}}\right)^2}\,dx=\\ 4\pi\int_{0}^{r}\sqrt{r^2-x^2}\frac{\sqrt{r^2-x^2+x^2}}{\sqrt{r^2-x^2}}\,dx= 4\pi r\int_{0}^{r}\,dx=4\pi rx\bigg|_{0}^{r}=4\pi r(r-0)=4\pi r^2.$$

• Thanks for the help! – Aspirant Feb 18 '19 at 17:49

You actually need to calculate $$2\int_{y\in [0,\,r]}2\pi y ds$$ with $$s$$ satisfying $$s^{\prime 2}=1+y^{\prime 2},\,f^\prime:=\frac{df}{dx}$$. This is because two circumferences separated by a distance $$dy$$ border a ribbon with a sloped edge of length $$ds$$.

• But shouldn’t $y= \sqrt{r^2-x^2}$ remedy that ? As it did , in case of using the disk-method for calculating the volume of the sphere ? – Aspirant Feb 18 '19 at 17:09
• @Aspirant The formula for $y$ doesn't magically change $ds$ into $dy$, but the volume formula doesn't use $ds$. I recommend you try the calculation I specified using $y^2=r^2-x^2\implies ydy=-xdx$. – J.G. Feb 18 '19 at 17:22