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\def\footer{\sl Quiz 11/22/91}

1. Let $R$ be the region bounded by 
$$y = x, \quad xy = 2, \quad x y^3 = 1$$
Rewrite the integral $\iint_R x^2 y^2\, dA$ as an iterated integral in the variables $u = x y$ and $v = x/y$ (do not evaluate).

\bigskip
{\sl Solution:}

Solve for $x$ and $y$ in terms of $u$ and $v$:
$$\displaylines{
u = x y \quad v = {x\over y}\cr
uv =  x y {x\over y} = x^2\quad {u\over v} = {xy \over x/y} = y^2\cr
x = (uv)^{1/2}\quad y = (u/v)^{1/2}\cr
}$$
Compute the Jacobian:
$$\eqalign{
{\partial(x,y)\over\partial(u,v)} &= \det \bmatrix{
{1\over2}(uv)^{-1/2}v& {1\over2}(uv)^{-1/2}u\cr
{1\over2}({u\over v})^{-1/2}{1\over v}&{1\over2}({u\over v})^{-1/2}(-{u\over v^2})\cr
}\cr
&= {1\over 4}(uv{u\over v})^{-1/2}(-{u\over v}) - {1\over 4}(uv {u\over v})^{-1/2}{u\over v}\cr
&= {1\over 4}{1\over u}(-{u\over v}) - {1\over 4}{1\over u}{u\over v}\cr
&= -{1\over 4 v} -{1\over 4 v} = -{1\over 2v}\cr
}$$ 
Transform the equations of $R$ into equations in $u$ and $v$:
$$\eqalign{
y = x &\to (uv)^{1/2} = (u/v)^{1/2}\to uv = u/v\cr
&\to v^2 = 1 \to v = 1\cr
x y = 2 &\to u = 2\cr
x y^3 = 1 &\to (xy) y^2 = 1 \to u (u/v) = 1\cr
&\to v = u^2\cr
}$$
\centerline{\epsfbox{solution.911122.1.eps}}
The new region is defined by
$$ 1 \le v \le u^2,\quad 1 \le u \le 2$$
$$\eqalign{
\iint_R x^2  y^2 \, dA &= \int_1^2\int_1^{u^2} (uv) (u/v) \left|-{1\over 2v}\right|\,dv\,du\cr
&= \int_1^2\int_1^{u^2}{u^2\over 2v}\,dv\,du\cr
}$$


Note: In this problem it is easier to compute the Jacobian `backwards':
$$\eqalign{
{\partial(u,v)\over\partial(x,y)} &= \det \bmatrix{
y& x\cr
{1\over y}&-{x\over y^2}\cr
}\cr
&= -{x\over y} -{x\over y} = -2{x\over y} = -2v\cr
{\partial(x,y)\over\partial(u,v)} &= \left[{\partial(u,v)\over\partial(x,y)}\right]^{-1} = -{1\over 2v}\cr
}$$

\vfill\hrule\bigskip
2. $\F(x,y) = x^2 y \i + y^3 z \j + z^4 x \k$.
$$\eqalign{
\div \F &= \d{}/dx(x^2 y) + \d{}/dy (y^3 z) + \d{}/dz(z^4 x)\cr
&= 2xy + 3y^2 z + 4z^3 x\cr
}$$

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