Surface area of a spherical dish

Question

How do I calculate the surface area of a spherical dish with a radius of curvature of 265 meters and a diameter of 305 meters? We can't use the normal formula 4(pi)r^2 because the sphere isn't a full or exactly half. How would you do it?

Answer 1

In the figure below, I've cut a sphere of radius $265$ along a latitude to obtain a spherical bowl of diameter $305$ (I'm just showing a cross-section). We can see a right triangle with hypotenuse $265$, and side opposite angle $\phi$ of length $152.5$. Then we can find $\phi = \arcsin\left(\frac{152.5}{265}\right) \approx 0.613 \text{ rad}$.

If you are familiar with surface integrals and spherical coordinates, the surface area of the bowl can now be computed as the integral of $1$ over the region $0 \leq \theta \leq 2\pi$, $0 \leq \phi \leq \arcsin\left(\frac{152.5}{265}\right)$, $\rho = 265$.

If you've been studying solids of revolution, then maybe you are familiar with the surface area formula $\int_a^b 2\pi x \sqrt{1+\left(\frac{dy}{dx}\right)^2} \ dx$ for the surface obtained by rotation $y=f(x)$, $a \leq x \leq b$, about the $y$-axis. The right half of the cross section of the bowl below is described by $f(x)=\sqrt{265^2-x^2}$, $0 \leq x \leq 152.5$ (assuming we center the sphere at the origin). So $$\frac{dy}{dx}= \frac{1}{2}(265^2-x^2)^{-1/2}\cdot -2x = \frac{x}{\sqrt{265^2-x^2}}$$ and $$\sqrt{1+\left(\frac{dy}{dx}\right)^2} = \sqrt{1+ \left(\frac{x}{\sqrt{265^2-x^2}}\right)^2}= \sqrt{1+ \frac{x^2}{265^2-x^2}}=\frac{265}{\sqrt{265^2-x^2}}.$$ So we end up with $$\int_0^{152.5}2\pi x\frac{265}{\sqrt{265^2-x^2}} \ dx$$ which can be solved by $u$-substitution with $u=265^2-x^2$.

This page might be useful to you: http://tutorial.math.lamar.edu/Classes/CalcII/SurfaceArea.aspx