Surface area of sphere derivation in polar co-ordinates query.

Question

Trying to work out the surface area of a sphere using polar co-ordinates but its going wrong.

Attempt:

Let $r=R\cos x$

If we rotate around the x axis:

$A=\int_{}^{} 2\pi y ds $

$A=\int_{}^{} 2\pi y \sqrt{r^2dx^2+dr^2}=\int_{}^{} 2\pi y \sqrt{r^2+\frac{dr^2}{dx^2}}dx$

$A=\int_{}^{} 2\pi r\sin x \sqrt{r^2+\frac{dr^2}{dx^2}}dx=\int_{}^{} 2\pi R\cos x\sin x \sqrt{R^2(\sin^2x+\cos^2x)}dx=\int_{}^{} \pi R^2\sin 2xdx$

Now I believe the limits should be $x_{upper}=\pi/2$ and $x_{lower}=0$ As the graph of r=Rcosx between these bounds looks like a semi circle, so rotating this around the x axis should produce a sphere.

But if I use these bounds: $A=\int_{0}^{\pi/2} \pi R^2\sin 2xdx=-\frac{\pi R^2}{2}\cos(\pi)+\frac{\pi R^2}{2}\cos(0)=\pi R^2$

Where have I gone wrong?

Answer 1

In your first equation $r$ is in error. It should be double that.

$$ r= 2R \cos x $$

In polar coordinates the equation of radius vector $r$ in a semi-circle centered on x-axis passing through the origin is a projection of diameter $2R$ as shown.

All else is correct. With

$$ R\rightarrow 2R,$$

you get $$ A = 4 \pi R^2 $$

Answer 2

Since $r^2dx^2+dr^2=R^2(\cos^2x+\sin^2x)dx^2$, with appropriate limits $A=2\pi R\int ydx$. Now we get to two mistakes. First, since you rotated a semicircle, not a quarter-circle the integration range should be from $0$ to $\pi$. Secondly, since $y^2+r^2=R^2$ we have $y=R\sin x$, not $y=r\sin x$. Thus $A=2\pi R^2\int_0^\pi\sin xdx=4\pi R^2$.