Section 9.4: 1, 2

And:

Let $B$ be the region bounded by the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a > 0$ and $b > 0$, and let $D$ be the closed disk $\bar B((0,0); 1)$. Show that

$\displaystyle{\int\int_B dA = ab\int\int_D dA}$,

and hence that the area of $B$ is $ab\pi$. Hint: Consider the transformation $T(u, v) = (au, bv)$.