22. Parametric Surfaces and Surface Integrals

d. Applications of Scalar Surface Integrals

1. Surface Area

Finding surface area of any given surface, \(S\), is straightforward now that we have found \(dS\). We integrate \(dS\) over the entire surface to find the total area. \[ A=\iint_S \,dS =\iint_S |\vec{N}|\,du\,dv \]

We need to know \(dS\) for spherical coordinates for a sphere of radius 2. Then we need the surface area of the piece of the sphere from \(\phi=0\) to \(\phi=30^\circ\).

The parametric equation for a sphere of radius \(2\) is \[ \vec R(\phi,\theta) =\left\langle 2\sin\phi\cos\theta,2\sin\phi\sin\theta,2\cos\phi\right\rangle \] From this equation we can find \(\vec{e}_\phi\) and \(\vec{e}_\theta\) (the radius \(\rho\) is not changing) \[ \begin{matrix} \quad \\ \vec{e}_\phi= \\ \vec{e}_\theta= \end{matrix} \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ \left\langle 2\cos\phi\cos\theta\right. & 2\cos\phi\sin\theta & \left. -2\sin\phi\right\rangle \\ \left\langle -2\sin\phi\sin\theta\right. & 2\sin\phi\cos\theta & \left. \quad0\quad\right\rangle \end{vmatrix} =\vec{N} \] Notice how we added the \(\hat{\imath}\), \(\hat{\jmath}\), \(\hat{k}\) and the vertical bars to compute \(\vec N\): \[\begin{aligned} \vec{N} &=\vec{e}_\theta\times\vec{e}_\phi \\ &=\hat{\imath}(0--4\sin^2\phi\cos\theta) -\hat{\jmath}(0-4\sin^2\phi\sin\theta) \\ &\quad+\hat{k}(4\sin\phi\cos\phi\cos^2\theta--4\sin\phi\cos\phi\sin^2\theta) \\ &=\left\langle 4\sin^2\phi\cos\theta,4\sin^2\phi\sin\theta,4\sin\phi\cos\phi\right\rangle \end{aligned}\]

The normal vector for a sphere must be radially inward or outward. In other words, it is a multiple of the position vector. In particular, the vector, \(\vec{N}\), in our example can be rewritten as: \[ \vec{N}=2\sin\phi\left\langle 2\sin\phi\cos\theta,2\sin\phi\sin\theta,2\cos\phi\right\rangle \] which is a multiple of the position vector. This is a useful mental check when integrating over the surface of a sphere.

To find \(dS\), we need the length of the normal: \[\begin{aligned} |\vec{N}| &=\sqrt{16\sin^4\phi\cos^2\theta+16\sin^4\phi\sin^2\theta+16\sin^2\phi\cos^2\phi} \\ &=\sqrt{16\sin^4\phi+16\sin^2\phi\cos^2\phi} =\sqrt{16\sin^2\phi} =4\sin\phi \end{aligned}\] Not coincidentally, this is the same thing you would get if you calculated the spherical Jacobian for volume, \(\rho^2\sin\phi\). Using the length of the normal, we have the scalar differential of surface area is: \[ dS=|\vec{N}|\,d\phi\,d\theta=4\sin\phi\,d\phi\,d\theta \] So we can now compute the integral for the area: \[\begin{aligned} A&=\iint_P \,dS =\iint_P |\vec{N}|\,d\phi\,d\theta =\int_0^{2\pi}\int_0^{\pi/6} 4\sin\phi\,d\phi\,d\theta \\ &=2\pi\left[-4\cos\phi\dfrac{}{}\right]_0^{\pi/6} =8\pi\left(-\cos\dfrac{\pi}{6}+\cos0\right) =8\pi\left(1-\,\dfrac{\sqrt{3}}{2}\right) \end{aligned}\]