a) Show that the formula for the surface area of a sphere with radius \(r\) is \(4 \pi r^{2}\). b) If a portion of the sphere is removed to form a spherical cap of height \(h\) then then show the curved surface area is \(2 \pi h r^{2} ?\)

Answer 1:

A = int dA

Explanation:

An area element on a sphere has constant radius r, and two angles. One is longitude \(\phi\), which varies from \(0\) to \(2π\). The other one is the angle with the vertical. To avoid counting twice, that angle only varies between \(0\) and \(π.\)

So the area element is \(d A=r d \theta r \sin \theta d \phi=r^{2} \sin \theta d \theta d \phi\) Integrated over the whole sphere gives

\(A=\int_{0}^{\pi} \sin \theta d \theta \int_{0}^{2 \pi} d \phi r^{2}=-\left.\cos (\theta)\right|_{0} ^{\pi} 2 \pi r^{2}=4 \pi r^{2}\)

In part b, \(\cos (\theta)\) varies between \(\frac{a}{r}\) and \(\frac{a}{r}\) which is such that \(b-a=h\)

Then \(A=\left.\cos \theta\right|_{b} ^{a} 2 \pi r^{2}=\left(\frac{b}{r}-\frac{a}{r}\right) 2 \pi r^{2}=\left(\frac{h}{r}\right) 2 \pi r^{2}=2 \pi r h\)

Note: Every derivation I found of this result uses cylindrical coordinates and is far more involved than this one. Can someone else check?

Answer 2:

\(\text { a) } A=4 \pi r^{2}\)

\(\text { b) } A=2 \pi h r^{2}\)

Explanation:

It is easier to use Spherical Coordinates, rather than Cylindrical or rectangular coordinates. This solution looks long because I have broken down every step, but it can be computer in just a few lines of calculation

With spherical coordinates, we can define a sphere of radius \(r\) by all coordinate points where \(0 \leq \phi \leq \pi\) (Where \(\phi\) is the angle measured down from the positive \(z\) -axis), and \(0 \leq \theta \leq 2 \pi\) (just the same as it would be polar coordinates), and \(\rho=r\))

The Jacobian for Spherical Coordinates is given by \(J=\rho^{2} \sin \phi\)

And so we can calculate the surface area of a sphere of radius \(r\) using a double integral: