How to calculate area of these overlapping rings?

Question

I want to calculate the surface area of what the brown paint stripes cover on the sphere. $r = 5$ cm.

Answer 1

In short, $A = 2\left(\frac{\theta}{360} \cdot 4 \pi r^2\right) - \frac{\theta}{360}\left(\frac{\theta}{360} \cdot 4 \pi r^2\right)$.

Really what's trying to be found is part of the surface area. The surface area of a sphere is $A = 4 \pi r^2$. To work that out you'd need to provide the radius.

But we're only interested in a fraction of the area, the thickness of the brown rings. To relate the thickness to the radius, imagine the sphere as just a circle with the same radius as the sphere (you can think of it as cutting right down the middle of the sphere). Then the thickness will be some part of the circumference. To find a fraction of the circumference, or in other words the length of an arc, you can do (assuming degrees) $\text{arc length} = \frac{\theta}{360} \pi d$. The same can be applied to the surface area of a sphere, giving $A = \frac{\theta}{360} \cdot 4 \pi r^2$.

This is done twice since there are two rings. However, there is an overlap. You can think of this similarly as the whole ring. We're only interested in a short part of the ring, more specifically a part of the ring that's the size of the width, in other words, we're interested in a fraction of the ring, the same fraction's the thickness. So we need to subtract $\frac{\theta}{360}$ times the ring's area from the total.

So in short, $A = 2\left(\frac{\theta}{360} \cdot 4 \pi r^2\right) - \frac{\theta}{360}\left(\frac{\theta}{360} \cdot 4 \pi r^2\right)$.

To demonstrate an example where the radius is 10 units and the ring is 5 units, first rearrange the arc length equation to find $\theta$:

\begin{align} \text{arc length} &= \frac{\theta}{360} \pi d \\ \theta &= 360\left(\frac{\text{arc length}}{\pi d}\right) \\ &= 360\left(\frac{5}{\pi (10 \div 2)}\right)\\ &= \frac{360}{\pi} ^\circ \end{align}

Then substitute:

\begin{align} A &= 2\left(\frac{\theta}{360} \cdot 4 \pi r^2\right) - \frac{\theta}{360}\left(\frac{\theta}{360} \cdot 4 \pi r^2\right)\\ &= 2\left(\frac{\frac{360}{\pi}}{360} \cdot 4 \pi 5^2\right) - \frac{\frac{360}{\pi}}{360}\left(\frac{\frac{360}{\pi}}{360} \cdot 4 \pi 5^2\right)\\ &= 168.1 \dots\\ &\approx 168 \text{ units}^2 \end{align}