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\centerline
{
\bf Solutions to the `QUIZ'' for Lecture 20
}
\bigskip
{\bf 1.} 
Find an equation for the tangent plane to the parametric surface
$$
x=v^2 \quad , \quad y=u+v \quad, \quad z=u^2 \quad ,
$$
at the point $(1,2,1)$. Simplify as much as you can!

{\bf Sol.} Here
$$
{\bf r}(t) = \langle v^2, u+v, u^2 \rangle
$$
Taking derivatives with respect to $u$ and $v$, we get
$$
{\bf r}_u= \langle 0, 1, 2u \rangle \quad,
$$
$$
{\bf r}_v= \langle 2v, 1, 0 \rangle \quad .
$$

Next, we have to find out what are $u$ and $v$ {\bf at}
the point $(1,2,1)$. We have to solve, for $u,v$:
$$
1=v^2, 2=u+v, 1=u^2 
$$
From the first equation $v=-1$ or $v=1$, from the last,
$u=-1$ or $u=1$, but to satisfy the second equation,
only $u=1$ and $v=1$ are OK. So we know that at the designated
point, $u=1,v=1$.

Plugging these above gives:
$$
{\bf r}_u= \langle 0, 1, 2 \rangle \quad,
$$
$$
{\bf r}_v= \langle 2, 1, 0 \rangle \quad .
$$

To find the normal, we take the cross-product
$$
{\bf n}=
\langle 0, 1, 2 \rangle  \times \langle 2, 1, 0 \rangle 
=\langle -2, 4, -2 \rangle \quad .
$$
(you do it!).

The equation of the tangent plane is
$$
\langle x-x_0 , y - y_0, z - z_0 \rangle \cdot  {\bf n}  =0 \quad,
$$
So, in this problem, it is
$$
\langle x-1 , y - 2, z - 1 \rangle \cdot  
\langle -2,4,-2 \rangle
  =0 \quad,
$$
that spells out to:
$$
(-2)(x-1)+4(y-2)+(-2)(z-1)=0 \quad .
$$
Dividing both sides by $-2$ and simplifying, we get
$$
x-2y+z=-2 \quad .
$$

{\bf Ans.} $x-2y+z=-2$ (type: Eq. of a plane).

{\bf Comments}: About \%40 got it perfectly, another \%20
got it correctly, but didn't completely simplify,
another \%20 did it the right way but messed up somewhere.
Some people did a very bad mistake, by not plugging in
$u=1, v=1$. You had to find what $u$ and $v$ are at the
designated point, and then plug-them-in. If you are not sure
how to find the $u$ and $v$ (like I did above), you should confess,
and stop right there. Leaving $u$ and $v$ in the answer is {\bf nonsense}!

{\bf 2.} 
Evaluate the surface integral
$$
\int\int_S  \, z \, dS \quad ,
$$ 
where $S$ is the triangular region
with vertices $(2,0,0),(0,2,0),(0,0,2)$.

{\bf Sol.} We first find the equation of the plane passing through
the three points. This turns out to be
$$
x+y+z=2 \quad .
$$
(in this easy case you can do it by ``inspection'' (adding
up the three coordinates always gives you $2$, in general
you would have to work hard, doing ${\bf n}={\bf AB} \times {\bf AC}$ etc.)

Expressing this plane in {\bf explicit} form, we have
$$
z=2-x-y \quad .
$$
The relevant formula is:
$$
\int\int_S f(x,y,z) dS = \int\int_D f(x,y,g(x,y)) \sqrt{ 1+ g_x^2+g_y^2} \, dx
\, dy \quad ,
$$
where $D$ is the projection of the region on the $xy$-plane.

Here $g(x,y)=2-x-y$, so $g_x=-1, g_y=-1$, and
$\sqrt{ 1+ g_x^2+g_y^2}=\sqrt{3}$. So
$$
\int\int_S  \, z \, dS =
\int\int_D  \, (2-x-y) \, \sqrt{3}  \quad .
$$ 
It still remains to find out the region $D$.
The plane $z=2-x-y$ meets the $xy$ plane (alias $z=0$) at
the line $x+y=2$. Since $x \geq 0, y \geq 2$
the region $D$ is
$$
D= \{ (x,y) \vert x \geq 0, y \geq 0, x+y \leq 2 \} \quad .
$$
A type I description is
$$
D=\{(x,y) \vert 0 \leq x \leq 2, 0 \leq y \leq 2-x \} \quad .
$$
So we get
$$
\int_0^2 \int_0^{2-x} \sqrt{3} \, dy \, dx \quad .
$$
The inner integral is
$$
\int_0^{2-x} \sqrt{3} \, dy =
\sqrt{3} y \Bigl \vert_{0}^{2-x}=
\sqrt{3} (2-x) \quad .
$$
The outer integral is:
$$
\int_0^2 \sqrt{3} (2-x) \, dx =
\sqrt{3} (2x-  \frac{x^2}{2}) \Bigl \vert_{0}^{2}=
2 \sqrt{3}  \quad .
$$
{\bf Ans.}: $2 \sqrt{3}$ (type: number).

{\bf Comments:} I really didn't allow enough time, 
so no one got it completely. Quite a few courageous people
almost got it, but only messed up in figuring out
$D$, and took it as $[0,2] \times [0,2]$.



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