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-[1/(x+y)]
1
x + y
correct
72
72
incorrect
at least one of the answers above is NOT correct.
(5 points)
Let $I = \iint_D (x^2 - y^2) \, dx \, dy$, where
$D = \{(x, y) : 2 \le xy \le 6, 0 \le x - y \le 3, x \ge 0, y \ge 0\}$
Show that the mapping $u = xy$, $v = x - y$ maps $D$ to the rectangle $R = [2, 6] \times [0, 3]$.
(a) Compute $\frac{\partial(x, y)}{\partial(u, v)}$ by first computing $\frac{\partial(u, v)}{\partial(x, y)}$.
(b) Use the Change of Variables Formula to show that $I$ is equal to the integral of $f(u, v) = v$ over $R$
and evaluate.
(a) $\frac{\partial(x,y)}{\partial(u,v)} = $
-(1/(x+y))
(b) $I = 72
