MvCalc67

From Example Problems
Jump to: navigation, search

\iint _{{R}}(x+y)e^{{x^{2}-y^{2}}}dA\,

where _{{R}}\, is the rectangle enclosed by the lines

x-y=0\,

x-y=2\,

x+y=0\,

x+y=3\,

For change of variables, Let

u=x-y\, and

v=x+y\,

So then by solving for x and y we get...

x={\frac  {1}{2}}(u+v)\, and

y={\frac  {1}{2}}(v-u)\,

0\leq u\leq 2\,

0\leq v\leq 3\,


\int _{{0}}^{{3}}\int _{{0}}^{{2}}(v)e^{{uv}}\cdot |J|dudv\, Where |J| is the Jacobian determinant


{\frac  {\partial (x,y)}{\partial (u,v)}}=\left[{\begin{matrix}{\frac  {1}{2}}&{\frac  {1}{2}}\\[6pt]-{\frac  {1}{2}}&{\frac  {1}{2}}\end{matrix}}\right]={\frac  {1}{2}}\,


{\frac  {1}{2}}\int _{{0}}^{{3}}\int _{{0}}^{{2}}(v)e^{{uv}}dudv\,


{\frac  {1}{2}}\int _{{0}}^{{3}}[e^{{uv}}]_{{0}}^{{2}}dv\,


{\frac  {1}{2}}\int _{{0}}^{{3}}(e^{{2v}}-1)dv\,


{\frac  {1}{2}}\int _{{0}}^{{3}}e^{{2v}}dv-{\frac  {1}{2}}\int _{{0}}^{{3}}dv\,


{\frac  {1}{2}}[{\frac  {1}{2}}e^{{2v}}]_{{0}}^{{3}}-{\frac  {1}{2}}[v]_{{0}}^{{3}}\,


{\frac  {1}{2}}[{\frac  {1}{2}}e^{{6}}-{\frac  {1}{2}}]-{\frac  {3}{2}}\,


{\frac  {1}{4}}e^{{6}}-{\frac  {7}{4}}\,


Hence..\iint _{{R}}(x+y)e^{{x^{2}-y^{2}}}dA={\frac  {1}{4}}(e^{{6}}-7)\,