Exercises — Parametric surfaces — tangent planes, surface area

${\mathbf{r}}_u$ means "differentiate each component with respect to $u$, holding $v$ fixed."

• $x=u\Rightarrow x_u=1$; $y=v\Rightarrow y_u=0$; $z=3u-2v\Rightarrow z_u=3$. ${\mathbf{r}}_u=⟨1,0,3⟩$ Same for $v$ (hold $u$ fixed):
• $x_v=0,y_v=1,z_v=-2$. ${\mathbf{r}}_v=⟨0,1,-2⟩$

Compute the cross product using the component rule $\mathbf{a}\times \mathbf{b}=⟨a_2{b}_3-a_3{b}_2,a_3{b}_1-a_1{b}_3,a_1{b}_2-a_2{b}_1⟩.$ With ${\mathbf{r}}_u=⟨1,0,3⟩$, ${\mathbf{r}}_v=⟨0,1,-2⟩$: ${\mathbf{r}}_u\times {\mathbf{r}}_v=⟨(0)(-2)-(3)(1),(3)(0)-(1)(-2),(1)(1)-(0)(0)⟩=⟨-3,2,1⟩.$ This is not the zero vector, so yes, the surface is smooth everywhere. (It is in fact a flat plane, which is as smooth as it gets.)

Point. $\mathbf{r}(2,1)=⟨2,1,2^2-1^2⟩=⟨2,1,3⟩$, so $P_0=(2,1,3)$.

Tangent arrows. ${\mathbf{r}}_u=⟨1,0,2u⟩=⟨1,0,4⟩,{\mathbf{r}}_v=⟨0,1,-2v⟩=⟨0,1,-2⟩.$

Normal = cross product: $\mathbf{n}={\mathbf{r}}_u\times {\mathbf{r}}_v=⟨(0)(-2)-(4)(1),(4)(0)-(1)(-2),(1)(1)-0⟩=⟨-4,2,1⟩.$

Plane. Point $P_0$ plus normal $⟨-4,2,1⟩$ gives $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$: $-4(x-2)+2(y-1)+1(z-3)=0⟹-4x+2y+z+3=0,$ or solved for $z$: $z=4x-2y-3$.

${\mathbf{r}}_u=⟨\cos v,\sin v,1⟩,{\mathbf{r}}_v=⟨-u\sin v,u\cos v,0⟩.$ Cross product: ${\mathbf{r}}_u\times {\mathbf{r}}_v=⟨(\sin v)(0)-(1)(u\cos v),(1)(-u\sin v)-(\cos v)(0),(\cos v)(u\cos v)-(\sin v)(-u\sin v)⟩$ $=⟨-u\cos v,-u\sin v,u{\cos }^{2}v+u{\sin }^{2}v⟩=⟨-u\cos v,-u\sin v,u⟩.$ Length: $|{\mathbf{r}}_u\times {\mathbf{r}}_v|=\sqrt{u^2{\cos }^{2}v+u^2{\sin }^{2}v+u^2}=\sqrt{u^2+u^2}=\sqrt{2}u.$ (We used ${\cos }^{2}v+{\sin }^{2}v=1$ twice — the workhorse identity for every trig-parametrized surface.)

${\mathbf{r}}_u=⟨2u,v,0⟩,{\mathbf{r}}_v=⟨0,u,2v⟩.$ ${\mathbf{r}}_u\times {\mathbf{r}}_v=⟨(v)(2v)-(0)(u),(0)(0)-(2u)(2v),(2u)(u)-(v)(0)⟩=⟨2v^2,-4uv,2u^2⟩.$ This is $0$ exactly when all three components vanish: $2v^2=0,-4uv=0,2u^2=0⟹u=0\text{and}v=0.$ Only the single parameter point $(u,v)=(0,0)$ fails, mapping to the surface point $(0,0,0)$.

Geometry: at $(0,0)$ both tangent arrows collapse to the zero vector (${\mathbf{r}}_u={\mathbf{r}}_v=0$). There is no parallelogram to speak of, so no plane is spanned — the parametrization pinches to a point (a degenerate patch). The surface may still look fine there; it's the map that is singular.

${\mathbf{s}}_u=⟨1,1,0⟩,{\mathbf{s}}_v=⟨1,-1,0⟩.$ ${\mathbf{s}}_u\times {\mathbf{s}}_v=⟨(1)(0)-(0)(-1),(0)(1)-(1)(0),(1)(-1)-(1)(1)⟩=⟨0,0,-2⟩,$ so $|{\mathbf{s}}_u\times {\mathbf{s}}_v|=2$.

Why area still matches: this factor of $2$ is the local area-stretch of B's map — it is precisely the Jacobian of the linear substitution $(x,y)=(u+v,u-v)$. B's parameter region $D_B$ is smaller by a factor of $2$ than A's unit square (because B's map squashes/rotates the $(u,v)$ grid onto the square). Integrating the stretch factor $2$ over that half-sized region gives $2\times \frac{1}{2}=1$: the same physical area A gets from ${\iint }_{[0,1{]}^2}1dA=1$. The stretch factor and the region size always compensate.

Use the graph parametrization $\mathbf{r}(x,y)=⟨x,y,x^2+y^2⟩$. With $f=x^2+y^2$: $f_x=2x,f_y=2y,|{\mathbf{r}}_x\times {\mathbf{r}}_y|=\sqrt{f_x^2+f_y^2+1}=\sqrt{4x^2+4y^2+1}.$ So $A={\iint }_{x^2+y^2\le 4}\sqrt{4x^2+4y^2+1}dA.$ The integrand depends only on $r^2=x^2+y^2$, so switch to polar coordinates ($x=r\cos \theta ,y=r\sin \theta$, area element $dA=rdrd\theta$ — this factor $r$ is itself a Jacobian, see Double Integrals): $A={\int }_0^{2\pi }{\int }_0^2\sqrt{4r^2+1}rdrd\theta .$ Inner integral: let $w=4r^2+1$, so $dw=8rdr\Rightarrow rdr=\frac{dw}{8}$. As $r:0\to 2$, $w:1\to 17$. ${\int }_0^2\sqrt{4r^2+1}rdr=\frac{1}{8}{\int }_1^{17}\sqrt{w}dw=\frac{1}{8}\cdot \frac{2}{3}{w}^{3/2}{|}_1^{17}=\frac{1}{12}(17^{3/2}-1).$ Multiply by ${\int }_0^{2\pi }d\theta =2\pi$: $\boxed{A=\frac{2\pi }{12}(17^{3/2}-1)=\frac{\pi }{6}(17\sqrt{17}-1)\approx 36.18}$

${\mathbf{r}}_u=⟨\cos v,\sin v,0⟩,{\mathbf{r}}_v=⟨-u\sin v,u\cos v,1⟩.$ ${\mathbf{r}}_u\times {\mathbf{r}}_v=⟨(\sin v)(1)-(0)(u\cos v),(0)(-u\sin v)-(\cos v)(1),u{\cos }^{2}v+u{\sin }^{2}v⟩=⟨\sin v,-\cos v,u⟩.$ $|{\mathbf{r}}_u\times {\mathbf{r}}_v|=\sqrt{{\sin }^{2}v+{\cos }^{2}v+u^2}=\sqrt{1+u^2}.$ $A={\int }_0^{\pi }{\int }_0^1\sqrt{1+u^2}dudv.$ The standard antiderivative $\int \sqrt{1+u^2}du=\frac{1}{2}(u\sqrt{1+u^2}+\mathrm{arcsinh}u)$ (this is the Arc Length integrand's cousin). Evaluate $0\to 1$: ${\int }_0^1\sqrt{1+u^2}du=\frac{1}{2}(\sqrt{2}+\ln (1+\sqrt{2})).$ Then multiply by ${\int }_0^{\pi }dv=\pi$: $\boxed{A=\frac{\pi }{2}(\sqrt{2}+\ln (1+\sqrt{2}))\approx 3.606}$

Compute the cross product. ${\mathbf{r}}_x=⟨1,0,f_x⟩,{\mathbf{r}}_y=⟨0,1,f_y⟩.$ ${\mathbf{r}}_x\times {\mathbf{r}}_y=⟨(0)(f_y)-(f_x)(1),(f_x)(0)-(1)(f_y),(1)(1)-(0)(0)⟩=⟨-f_x,-f_y,1⟩.$ Length. $|{\mathbf{r}}_x\times {\mathbf{r}}_y|=\sqrt{(-f_x{)}^2+(-f_y{)}^2+1^2}=\sqrt{f_x^2+f_y^2+1}.$ The inequality. Since $f_x^2+f_y^2\ge 0$ always, the integrand satisfies $\sqrt{f_x^2+f_y^2+1}\ge \sqrt{0+0+1}=1,$ with equality only where $f_x=f_y=0$ (a locally flat, horizontal spot). Integrating over $D$: $A(S)={\iint }_D\sqrt{f_x^2+f_y^2+1}dA\ge {\iint }_{D}1dA=\text{area}(D).$ If $f$ is non-flat, it has a slope on a region of positive area, so the inequality is strict there and $A(S)>\text{area}(D)$: tilting a surface can only add area — the flat shadow is always the smallest. (The $z$-component of ${\mathbf{r}}_x\times {\mathbf{r}}_y$ being $1$ is exactly why: it says the normal always "leans back" toward vertical, and length $\ge$ any single component.)

Magnitudes agree because $|d\mathbf{S}|=|({\mathbf{r}}_u\times {\mathbf{r}}_v)dA|=|{\mathbf{r}}_u\times {\mathbf{r}}_v|dA=dS$ (the scalar $dA>0$ pulls straight out of the magnitude).

• Area measures how much surface there is — a size, no direction — so it uses the length $|{\mathbf{r}}_u\times {\mathbf{r}}_v|$, the size of each little parallelogram.
• Flux measures how much of a vector field passes through the surface, which needs a direction to say "through which way." So it keeps the full vector ${\mathbf{r}}_u\times {\mathbf{r}}_v$, whose direction is the normal and whose length is still the patch area. One quantity, two uses: length for how-much, full vector for through-which-way.