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\centerline
{
\bf Solution to the ``QUIZ'' for Lecture 22
}
\bigskip

Evaluate the surface
integral $\int\int_S {\bf F} \cdot d {\bf S}$ for
the given vector field ${\bf F}$ and oriented
surface $S$.
$$
{\bf F}(x,y,z)= 
\langle xy \, , \, yz \, , \, zx \, \rangle \quad,
$$
and $S$ is the part of the paraboloid $z=1-x^2-y^2$ that lies
above the square $0 \leq x \leq 1 \, , \, 0 \leq y \leq 1$
and has upward orientation.


{\bf Solution}

The easiest way is to use the formula for {\bf explicitly}-defined surfaces
$z=g(x,y)$, and a vector field ${\bf F}=\langle P,Q,R \rangle$:
$$
\int\int_S {\bf F} \cdot d{\bf S}=
\int\int_D (
-P {{\partial g} \over {\partial x}}
-Q {{\partial g} \over {\partial y}}
+R) \, dA \quad .
$$
where $D$ is the ``floor'', in this problem $0 \leq x,y \leq 1$.

In this problem $P=xy,Q=yz,R=zx$, $g(x,y)=1-x^2-y^2$, so we have
$$
\int\int_D (-xy(-2x) -yz(-2y)+xz) dA 
=\int\int_D (2x^2y +2y^2z+xz) dA \quad .
$$
In addition we {\bf must} replace $z$ by $g(x,y)$ ($1-x^2-y^2$ in this
problem).
So we have
$$
\int\int_D (2x^2y +2y^2z+xz) dA \, = \, 
\int\int_D (2x^2y +2y^2(1-x^2-y^2)+x(1-x^2-y^2)) \, dA \quad .
$$
Doing the tedius algebra we get
$$
\int_0^1 \int_0^1 (2x^2y +2y^2-2y^2x^2-2y^4+x-x^3-xy^2) \, dx \, dy \quad .
$$

Now you do the inside integral, getting someting in $y$, and then
the outside integral, and it comes out to $\frac{83}{180}$.

{\bf Ans.} $\frac{83}{180}$ (type number).

{\bf Comments:} Most people set-it up correctly, but only
a few courageous souls finished it up to the bitter end.




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