Question

Use the given transformation to evaluate the given integral, where R is the parallelogram with vertices (-2, 6), (2, -6), (5,-3), and (1,9). L = SUR(16.+12y) dA; r = {(u +v), y=(v – 3u) L =

Answer 1

$\small \text{Consider the integral}\\ \int \int_R (16x+12y) dA\\ R: \; \text{is the trapezoidal region with vertices } (-2,6),(2,-6),(5,-3),(1,9)\\ \text{Let us assume that The substitution :}\\ x=\frac{1}{4}(u+v),\; \; y=\frac{1}{4}(v-3u)\\ j=\frac{\partial (x,y)}{\partial (u,v)}=\begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix}=\begin{vmatrix} \frac{1}{4} & \frac{1}{4}\\ -\frac{3}{4}& \frac{1}{4} \end{vmatrix}=\frac{1}{16}+\frac{3}{16}=\frac{4}{16}=\frac{1}{4}\\ x=\frac{1}{4}(u+v)\Rightarrow u+v=4x\\ y=\frac{1}{4}(v-3u)\Rightarrow v-3u=4y\\ \text{Solving these equations we get}\\$

$\small u=x-y,\: v=3x+y\\ \text{So, the boundaries of R transform into u-v plane is as follows}\\ \text{The point (-2,6) in x-y plane is transform to (-8,0) in u-v plane }\\ \text{The point (2,-6) in x-y plane is transform to (8,0) in u-v plane }\\ \text{The point (5,-3) in x-y plane is transform to (8,12) in u-v plane }\\ \text{The point (1,9) in x-y plane is transform to (-8,12) in u-v plane }\\ \text{Therefore, the integral become as follows}\\ \int \int _{\mathbb{R}}f(x,y)dy\: dx=\int \int _{\mathbb{S}}f(g(u,v),h(u,v))\left |j \right |du\: dv\\ L=\int \int_R (16x+12y) dA=\int_{v=0}^{12}\int_{u=-8}^{8}\left [ 16\cdot \frac{1}{4}(u+v)+12\cdot \frac{1}{4}(v-3u) \right ]\frac{1}{4}du\: dv\\ =\int_{v=0}^{12}\int_{u=-8}^{8}\left [ 4(u+v)+3(v-3u) \right ]\frac{1}{4}du\: dv\\ =\int_{v=0}^{12}\int_{u=-8}^{8}\left [ 4u+4v+3v-9u \right ]\frac{1}{4}du\: dv\\ =\int_{v=0}^{12}\int_{u=-8}^{8}(7v-5u)\frac{1}{4}du\: dv\\ =\int _0^{12}28vdv\\ =\left ( 14v^2 \right )_0^{12}\\ =2016-0\\ L=2016$