Math 230-2: Kursobjekt

1. First we parametrize $\mathcal{S}\text{.}$ Observe that intersecting $\mathcal{S}$ with the plane $z=z_0$ yields a circle centered at $(0,0,z_0)$ of radius $R=\cos z_0\text{,}$ which we can parametrize as

   $$(R\cos \theta ,R\sin \theta ,z_0)=(\cos z_0\cos \theta ,\cos z_0\sin \theta ,z_0)\text{.}$$

   Letting $z$ vary between $-\pi /2$ and $\pi /2\text{,}$ we get the parametrization

   $$\begin{array}{rl}\mathbf{r}(\theta ,z)& =(\cos z\cos \theta ,\cos z\sin \theta ,z)\\ \mathcal{R}:0& \le \theta \le 2\pi ,-\pi /2\le z\le \pi /2\text{.}\end{array}$$

2. Assemble the necessary ingredients:

   $$\begin{array}{rl}{\mathbf{r}}_{\theta }& =⟨-\cos z\sin \theta ,\cos z\cos \theta ,0⟩\\ {\mathbf{r}}_z& =⟨-\sin z\cos \theta ,-\sin z\sin \theta ,1⟩\\ {\mathbf{r}}_{\theta }\times {\mathbf{r}}_z& =⟨\cos z\cos \theta ,\cos z\sin \theta ,\sin z\cos z{\sin }^2\theta +\sin z\cos z{\cos }^2\theta ⟩\\ & =⟨\cos z\cos \theta ,\cos z\sin \theta ,\sin z\cos z⟩\\ {|{\mathbf{r}}_{\theta }\times {\mathbf{r}}_z|}_{}& =\sqrt{{\cos }^2(1+{\sin }^{2}z)}\\ & ={|\cos z|}_{}\sqrt{1+{\sin }^{2}z}\\ & =\cos z\sqrt{1+{\sin }^{2}z}\text{,}\end{array}$$

   where the last equality follows since $\cos z\ge 0$ for $z\in [-\pi /2,\pi /2]\text{.}$

3. According to Interpretation 2.6.5, integrating a mass density function over $\mathcal{S}$ gives us the total mass of $\mathcal{S}\text{.}$ Thus

   $$\begin{array}{rl}\mathrm{mass}\mathcal{S}& =\underset{\mathcal{S}}{\iint }f(x,y,z)d\sigma \\ & =\underset{\mathcal{R}}{\iint }f(\mathbf{r}(\theta ,z)){|{\mathbf{r}}_{\theta }\times {\mathbf{r}}_z|}_{}dA\\ & ={\int }_0^{2\pi }{\int }_{-\pi /2}^{\pi /2}\cos z\sqrt{1-{\cos }^{2}z{\cos }^2\theta -{\cos }^{2}z{\sin }^2\theta }\sqrt{1+{\sin }^{2}z}dzd\theta \\ & =2\pi {\int }_{-\pi /2}^{\pi /2}\cos z\sqrt{{\sin }^{2}z(1+{\sin }^{2}z)}dz\\ & =2\pi {\int }_{-\pi /2}^{\pi /2}{|\sin z|}_{}\cos z\sqrt{1+{\sin }^{2}z}dz\\ & =4\pi {\int }_0^{\pi /2}\sin z\cos z\sqrt{1+{\sin }^{2}z}dz,({|\sin z|}_{}\text{ even})\\ & =4\pi (\frac{1}{3}(1+{\sin }^{2}z{)}^{3/2}){|}_0^{\pi /2}\\ & =\frac{4\pi }{3}(2^{3/2}-1)\mathrm{grams}\text{.}\end{array}$$

We have $\mathcal{T}=T_1\cup T_2\cup T_3\cup T_4\text{,}$ where the $T_k$ are the four triangular faces of $\mathcal{T}\text{,}$ and thus

Let $T_1,T_2$ be the triangular faces lying in the coordinate planes $x=0$ and $y=0\text{,}$ respectively. Since $f(x,y,z)=0$ for all points in these two surfaces, the corresponding surface integrals ${\iint }_{T_k}fd\sigma \text{,}$ $k=1,2\text{,}$ are equal to zero. Thus we need only compute the integrals of $f$ over $T_3$ and $T_4\text{.}$

• Integral over $T_3$.

  Let $T_3$ be the face of the tetrahedron in the $xy$-plane. We parametrize as

  $$\begin{array}{rl}\mathbf{r}(x,y)& =(x,y,0)\\ {\mathcal{R}}_1:0& \le x\le 1,0\le y\le 1-x\text{.}\end{array}$$

  A simple computation shows ${|{\mathbf{r}}_x\times {\mathbf{r}}_y|}_{}=1\text{.}$ Thus

  $$\begin{array}{rl}\underset{T_1}{\iint }fd\sigma & ={\int }_0^1{\int }_0^{1-x}xydydx\\ & =\frac{1}{2}{\int }_0^{1}x(1-x{)}^{2}dx\\ & =-\frac{1}{2}{\int }_1^0(1-u)u^{2}du& (u=1-x,du=-dx)\\ & =\frac{1}{2}(1/3-1/4)=\frac{1}{24}\end{array}$$
• Integral over $T_4$.

  Let $T_4$ be triangle with vertices $(1,0,0),(1,0,0),(0,1,0)\text{.}$ This is the region of the plane $z=x+y$ lying over the planar triangle with vertices $(0,0,0),(1,0,0),(0,1,0)\text{.}$ Since $z$ is expressed as a function of $x,y$ we parametrize easily as

  $$\begin{array}{rl}\mathbf{s}(x,y)& =(x,y,x+y)\\ {\mathcal{R}}_2:0& \le x\le 1,0\le y\le 1-x\text{.}\end{array}$$

  Again, a straightforward computation shows ${|{\mathbf{s}}_x\times {\mathbf{s}}_y|}_{}=\sqrt{3}\text{,}$ and thus

  $$\begin{array}{rlr}\underset{T_4}{\iint }fd\sigma & =\underset{{\mathcal{R}}_2}{\iint }f(\mathbf{s}(x,y))\sqrt{3}dA& \\ & =\sqrt{3}{\int }_0^1{\int }_0^{1-x}xydydx\\ & =\frac{\sqrt{3}}{24}\text{,}\end{array}$$

  where we use our work from above for the last equality.

We use the usual parametrization of the sphere of radius 2:

Given a point $P=(x,y,z)\in \mathcal{S}$ the vector $⟨x,y,z⟩$ is an outward pointing normal vector of magnitude $\sqrt{x^2+y^2+z^2}=2\text{.}$ Thus the function $\mathbf{n}(x,y,z)=\frac{1}{2}⟨x,y,z⟩$ is the outward orientation on $\mathcal{S}\text{.}$ It follows that

and thus that

The last step above is done using the identity ${\sin }^2\theta =\frac{1}{2}(1-\cos 2\theta )$ and the fact that ${\sin }^3\varphi =\sin \varphi (1-{\cos }^2\varphi )=\sin \varphi -\sin \varphi {\cos }^2\varphi$ has antiderivative $-\cos \varphi +\frac{1}{3}{\cos }^3\varphi \text{.}$

Note that in this case we computed the surface integral by (a) identifying the outward orientation $\mathbf{n}$ on $\mathcal{S}$ geometrically (as opposed to using ${\mathbf{r}}_{\varphi }\times {\mathbf{r}}_{\theta }$ or its opposite), (b) computing the scalar function $\mathbf{F}(x,y,z)\cdot \mathbf{n}(x,y,z)\text{,}$ and (c) integrating with respect to $d\sigma =|{\mathbf{r}}_{\varphi }\times {\mathbf{r}}_{\theta }|=4\sin \varphi \text{.}$ This method works in this case because we could easily identify the orientation function $\mathbf{n}(x,y,z)\text{.}$ When this is not easily done, resort to the procedure indicated in Theorem 2.6.15. See Example 2.6.16 for an illustration of this technique.

• Parametrize. Using cylindrical coordinates, we see that $\mathcal{S}$ has parametrization

  $$\begin{array}{rl}\mathbf{s}(r,\theta )& =(r\cos \theta ,r\sin \theta ,r)\\ \mathcal{R}:0& \le \theta \le 2\pi ,0\le r\le 4\text{.}\end{array}$$

• Ingredients.

  $$\begin{array}{rl}{\mathbf{s}}_r& =⟨\cos \theta ,\sin \theta ,1⟩\\ {\mathbf{s}}_{\theta }& =⟨-r\sin \theta ,r\cos \theta ,0⟩\\ {\mathbf{s}}_r\times {\mathbf{s}}_{\theta }& =⟨-r\cos \theta ,-r\sin \theta ,r⟩\text{.}\end{array}$$

  Observe that since $r\ge 0\text{,}$ the normal vector ${\mathbf{s}}_r\times {\mathbf{s}}_{\theta }$ points upward, and hence toward the $z$-axis. Thus the outward orientation $\mathbf{n}$ comes from ${\mathbf{s}}_{\theta }\times {\mathbf{s}}_r\text{.}$

• Now compute

  $$\begin{array}{rl}{\iint }_{\mathcal{S}}\mathbf{F}\cdot \mathbf{n}d\sigma & ={\iint }_{\mathcal{R}}\mathbf{F}(r\cos \theta ,r\sin \theta ,r)\cdot ⟨r\cos \theta ,r\sin \theta ,-r⟩dA\\ & ={\iint }_{\mathcal{R}}⟨r^2\cos \theta \sin \theta ,0,-r⟩\cdot ⟨r\cos \theta ,r\sin \theta ,-r⟩dA\\ & ={\int }_0^{2\pi }{\int }_0^4{r}^3{\cos }^2\theta \sin \theta +r^{2}drd\theta \\ & =\frac{128}{3}\pi \text{.}\end{array}$$