22. Parametric Surfaces and Surface Integrals

d. Applications of Scalar Surface Integrals

4. Mass

The total mass is always the integral of the density, \(\delta\), over the region. In the case of parametric surfaces, the density is given in units of \(\dfrac{\text{mass}}{\text{area}}\) and the mass is then \[ M=\iint_S \delta\,dS =\iint_S \delta(\vec R(u,v))\,|\vec{N}|\,du\,dv \]

The sphere is parametrized by: \[ \vec R(\phi,\theta) =\left\langle 2\sin\phi\cos\theta,2\sin\phi\sin\theta,2\cos\phi\right\rangle \] In these coordinates, the density is \(\delta=z^2-3=4\cos^2\phi-3\). On a previous page, we found that \(|\vec{N}|=4\sin\phi\). So the total mass is: \[\begin{aligned} M&=\iint_P \delta\,dS =\iint_P \delta(\theta,\phi)|\vec{N}|\,d\phi\,d\theta \\ &=\int_0^{2\pi}\int_0^{\pi/6} (4\cos^2\phi-3)4\sin\phi\,d\phi\,d\theta \\ &=8\pi\int_0^{\pi/6} (4\cos^2\phi-3)\sin\phi\,d\phi \\ \end{aligned}\]

Here we use the substitution \(u=\cos\phi\). So \(du=-\sin\phi\,d\phi\). Therefore, the mass becomes: \[\begin{aligned} M&=-8\pi\int_1^{\sqrt{3}/2} (4u^2-3)\,du =-8\pi\left[4\dfrac{u^3}{3}-3u\right]_1^{\sqrt{3}/2} \\ &=-8\pi\left[\left(4\dfrac{3\sqrt{3}}{3\cdot8}-\,\dfrac{3\sqrt{3}}{2}\right) -\left(\dfrac{4}{3}-3\right)\right] \\ &=8\pi\sqrt{3}-\,\dfrac{40}{3}\pi \\ \end{aligned}\]