# Show that the formula for the surface area of a sphere with radius

Surface Area of a Sphere | Lecture 40 | Vector Calculus for Engineers
Surface Area of a Sphere | Lecture 40 | Vector Calculus for Engineers

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 $$2 \pi h r^{2} ?$$

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?

$$\text { a) } A=4 \pi r^{2}$$
$$\text { b) } A=2 \pi h r^{2}$$

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:

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