6.6 Surface integrals  (Page 8/27)

Notice that S is not smooth but is piecewise smooth; S can be written as the union of its base $S_1$ and its spherical top $S_2,$ and both $S_1$ and $S_2$ are smooth. Therefore, to calculate ${\displaystyle {\iint }_S{z}^{2}dS},$ we write this integral as ${\displaystyle {\iint }_{S_1}{z}^{2}dS}+{\displaystyle {\iint }_{S_2}{z}^{2}dS}$ and we calculate integrals ${\displaystyle {\iint }_{S_1}{z}^{2}dS}$ and ${\displaystyle {\iint }_{S_2}{z}^{2}dS}.$

First, we calculate ${\displaystyle {\iint }_{S_1}{z}^{2}dS}.$ To calculate this integral we need a parameterization of $S_1.$ This surface is a disk in plane $z=1$ centered at $(0,0,1).$ To parameterize this disk, we need to know its radius. Since the disk is formed where plane $z=1$ intersects sphere $x^2+y^2+z^2=4,$ we can substitute $z=1$ into equation $x^2+y^2+z^2=4\text{:}$

Therefore, the radius of the disk is $\sqrt{3}$ and a parameterization of $S_1$ is $\text{r}(u,v)=⟨u\text{cos}v,u\text{sin}v,1⟩,0\le u\le \sqrt{3},0\le v\le 2\pi .$ The tangent vectors are ${\text{t}}_u=⟨\text{cos}v,\text{sin}v,0⟩$ and ${\text{t}}_v=⟨\text{−}u\text{sin}v,u\mathrm{co}\mathrm{s}v,0⟩,$ and thus

The magnitude of this vector is u . Therefore,

Now we calculate ${\displaystyle {\iint }_{S_2}dS}.$ To calculate this integral, we need a parameterization of $S_2.$ The parameterization of full sphere $x^2+y^2+z^2=4$ is

Since we are only taking the piece of the sphere on or above plane $z=1,$ we have to restrict the domain of $\varphi .$ To see how far this angle sweeps, notice that the angle can be located in a right triangle, as shown in [link] (the $\sqrt{3}$ comes from the fact that the base of S is a disk with radius $\sqrt{3}).$ Therefore, the tangent of $\varphi$ is $\sqrt{3},$ which implies that $\varphi$ is $\pi \text{/}6.$ We now have a parameterization of $S_2\text{:}$

The tangent vectors are

and thus

The magnitude of this vector is

Therefore,

Since ${\displaystyle {\iint }_S{z}^{2}dS}={\displaystyle {\iint }_{S_1}{z}^{2}dS}+{\displaystyle {\iint }_{S_2}{z}^{2}dS},$ we have ${\displaystyle {\iint }_S{z}^{2}dS}=\left(2\pi -4\right)\sqrt{3}+\frac{32\pi }{3}.$