Volumes + Surface Areas of Sphere with Calculus: Paradox?

Question

I have an interesting observation, what appears to be an apparant contradiction.

We calculate volume of sphere, we get it as $$\int_{-R}^R \pi (\sqrt{R^2-y^2})^2 dy$$ because we see it as, circular cylinder with height dy.

We calculate surface area of sphere, we get it as $$\int_{-R}^R 2*\pi \sqrt{R^2-y^2} dy$$ because we see it as, circular cylinder outside with height dy.

First integral, well I guess it is right. $\frac{4}{3} \pi R^3$. No problem there. But then, you look the second integral, you get, $\pi^2 R^2$! What is discrepancy here, and what is wrong.

Answer 1

It is a good observation, showing that one has to be very careful when dealing with integral approximations via Riemann sums. It is easy to do a mistake here. To better visualize what's going on, let's look at an example in $\mathbb{R}^2$ where the "volume" is area and the "surface area" is length, i.e. everything has one dimension less.

When we introduce a Riemann integral we take piecewise constant approximations of $f(x)$ and approximate the integral by the sum of all small rectangles. (In your case it is circular cylinders in the first integral). So the grey area on the picture is close to the area under the graph.

If we would like to calculate the graph lenght instead, we need another kind of approximations - by piecewise linear curve (the blue one on the second picture) - so that the red length is approximated by the sum of all blue lengths $\Delta L$.

If in the second case we mistakenly take a piece-wise constant approximation as in the first case (the horisontal green lengths $\Delta x$) then we get the sum of all such $\Delta x$ to be exactly the length of the interval $[a,b]$, which is the same for all curves, and, of course, has nothing to do with the red length at all.

Precisely the same thing happens in your second integral approximation of the surface area. You can visualize the volume and the surface area by rotating the area and the length in the example above.

Answer 2

Now Remember the formula for the Volume/Surface Area of Solids of Revolution. Consider I have a function y = f(x) and I revolve it around the x-axis in this case between a and b. Then the volume of the 3 dimensional Shape is V = $\pi$ $\int_{a}^b y^2 dx $ And The Surface Area is $2\pi$ $\int_{a}^b y \sqrt{1+(\frac{dy}{dx})^{2} } dx$.

$1.$ Volume of the Sphere

Consider the Equation of the Circle Centered at the Origin with Radius r :- $x^{2}+ y^{2} =r^{2} $ . Subtract $x^{2}$ from both sides and you get $y^{2}$ = $r^{2}- x^2$ and Taking the Positive Square root on both sides we get $y = \sqrt{r^2-x^2}$ . The bounds will be from x = -r to x = r. Now then substitute every thing in the Formula for the Volume of Solids of Revolution $\pi$ [$\int_{-r}^r r^2-x^2 dx]$. Then $\pi$$(\int_{-r}^r r^2 dx$ -$\int_{-r}^r x^2 dx)$, then you get $\pi$ $(2r^3$ - $\frac{2r^3}{3})$ , doing some simple Algebra and rearranging all the terms and then you will get $\frac{4}{3}\pi r^3.$

$2.$ Surface Area Of the Sphere

Consider the same Equation $y = \sqrt{r^2-x^2} $ Now you compute $\frac{dy}{dx}$ which is $\frac{-x}{\sqrt{r^2-x^2}}.$ Now you substitute all the values in the second Surface Area formula and then you get $2\pi$ $\int_{-r}^r \sqrt{r^2-x^2} \sqrt{1+\frac{x^2}{r^2-x^2} } dx$ Then after simple algebra then this will evaluate to a very very simple integral $2\pi$ $\int_{-r}^r r dx $ And this you can do it yourself and then you will get the Answer $4\pi r^2.$

Thanks.