Derive the surface area of a sphere using thin rings with the natural coordinates for a sphere!

Rectangular coordinates are ill-suited to describing an object with spherical symmetry, so in this video we apply the natural coordinates for a sphere for deriving the surface area of a sphere using thin rings. We slice the sphere into thin ribbons and set up the integral for surface area of a sphere as a theta integral, where theta is the angular position of each ring, or ribbon. We start by labeling the dimensions of each ribbon, so we begin with the radius. It's critical to see that the radius of the ribbons is changing, from very small at the poles to maximum at the equator of the sphere. We find this variable radius as Rcos(theta) by using right triangle trigonometry. Next, we get the thickness of each ribbon, which is given by using the old formula for arc length and angle: s=R*theta, but this time it's an infinitesimal arc and the angle subtending the edge of the ribbon is also infinitesimal. So we get ds=R*d(theta). Now we cut and unroll the ring to produce a thin rectangle. The length is the circumference of the original rolled up ribbon, and the thickness is ds=R*d(theta). This allows us to quickly state the area contribution of the ribbon as dA=2pi*R^2*cos(theta)d(theta). Because the area contribution dA is phrased in terms of a single variable, we're ready to add up the contributions using an integral, so we integrate dA as theta goes from -pi/2 to pi/2. We simplify a little by using the parity of the cosine function to cut the integration interval in half. Finally, we find the antiderivative and evaluate across the limits of integration to obtain the familiar formula for the surface area of a sphere: 4pi*R^2.