Thm 5.3: Suppose T : R^{2} → R^{2} ∈ C^{1} T (u, v) = (T_{1}(u, v), T_{2}(u, v)) = (x(u, v), y(u, v)) If D^{∗} ⊂ R^{2} and if f : T (D^{∗}) → R is integrable, then ∫ ∫_{T (D^{∗})} f(x, y)dxdy =

∫ ∫_{D^{∗}} f((x(u, v), y(u, v)))abs(det(DT (u, v)))dudv Ex: Evaluate ∫ ∫_{T (D^{∗})} xydxdy where T (D^{∗}) = the parallelogram where the vectors (4, 1) and (2, 3) form two of its sides.

Find D^{∗} and T : R^{2} → R^{2} such that T (D^{∗}) is the parallelogram described above. Note T is a linear transformation.

T : R^{2} → R^{2}, T (u, v) = ( \begin{matrix}a & b \\ c & d \end{matrix} ) ( \begin{matrix}u \\ v \end{matrix} )

T (1, 0) = ( \begin{matrix}a & b \\ c & d \end{matrix} ) ( \begin{matrix}1 \\ 0 \end{matrix} ) = ( \begin{matrix}4 \\ 1 \end{matrix} )

T (0, 1) = ( \begin{matrix}a & b \\ c & d \end{matrix} ) ( \begin{matrix}0 \\ 1 \end{matrix} ) = ( \begin{matrix}2 \\ 3 \end{matrix} )

Hence T : R^{2} → R^{2},

T (u, v) = ( \begin{matrix}4 & 2 \\ 1 & 3 \end{matrix} ) ( \begin{matrix}u \\ v \end{matrix} ) = ( \begin{matrix}4u + 2v \\ u + 3v \end{matrix} ) = ( \begin{matrix}x(u, v) \\ y(u, v) \end{matrix} )

Let x = 4u + 2v and y = u + 3v

DT = ( \begin{matrix}4 & 2 \\ 1 & 3 \end{matrix} ) and abs(det ( \begin{matrix}4 & 2 \\ 1 & 3 \end{matrix} )) = 12 − 2 = 10

∫ ∫_{T (D^{∗})}(xy)dxdy = ∫ ∫_{D^{∗}} f((x(u, v), y(u, v)))abs(det(DT (u, v)))dudv

= ∫ ∫_{D^{∗}}(4u + 2v)(u + 3v)(10)dudv

= 10 ∫_{0}^{1} ∫_{0}^{1}(4u^{2} + 14uv + 6v^{2})dudv

= 10 ∫_{0}^{1}(\frac{4}{3}u^{3} + 7u^{2}v + 6uv^{2}|_{0}^{1}dv

= 10 ∫_{0}^{1}(\frac{4}{3} + 7v + 6v^{2})dv

= 10(\frac{4}{3}v + \frac{7}{2}v^{2} + 2v^{3})|_{0}^{1}

= 10(\frac{4}{3} + \frac{7}{2} + 2)