Question

Use the given transformation to evaluate the integral.

\(\iint_{\mathsf{R}}{{}}\left( {{\mathsf{x}}^{\mathsf{2}}}\mathsf{-xy+}{{\mathsf{y}}^{\mathsf{2}}} \right)\mathsf{dA}\)whereis the region bounded by the ellipse \({{\mathsf{x}}^{\mathsf{2}}}\mathsf{-xy+}{{\mathsf{y}}^{\mathsf{2}}}\mathsf{=2;x=}\sqrt{\mathsf{2}}\mathsf{u-}\sqrt{\mathsf{2/3}}\mathsf{v,y=}\sqrt{\mathsf{2}}\mathsf{u+}\sqrt{\mathsf{2/3}}\mathsf{v}\)

Short answer

The required integral is \(\frac{{4\sqrt 3 \pi }}{3}\).

Step-by-step solution

Formula

Formula for the Jacobian

\(\frac{{\partial (x,y)}}{{\partial (u,v)}} = \left| {\begin{array}{*{20}{l}}{\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}\\{\partial y}&{\partial y}\end{array}} \right|\)

Find partial derivatives and using Jacobian formula

\(\iint_{R}{{}}\left( {{x}^{2}}-xy+{{y}^{2}} \right)dA\)

We need to figure out what happens to the region when the given transformation is applied. When we make the substitution \(x=\sqrt{2}u-\sqrt{2/3}v\)

and \(y=\sqrt{2}u+\sqrt{2/3}v\),

we get \(2{{u}^{2}}+2{{v}^{2}}\) The ellipse \(({{x}^{2}}-xy+{{y}^{2}})\) is mapped to the unit circle \(({{u}^{2}}+{{v}^{2}}=1)\).

Jacobian = \(|{{x}_{u}}{{x}_{v}}{{y}_{u}}{{y}_{v}}|\)

Jacobian = \(|\sqrt{2}-\sqrt{2/3}\sqrt{2}\sqrt{2/3}|\) = \(\frac{{4\sqrt 3 \pi }}{3}\).

Therefore, our integral becomes

\(\iint_{D}{{}}\left( 2{{u}^{2}}+2{{v}^{2}} \right)\frac{4\sqrt{3}}{3}dA\)

Where D is the region \({{u}^{2}}+{{v}^{2}}\le 1\)

Simplify the integral equation

Solve this integral,

\(u = r\cos \theta ,v = r\sin \theta ,0 < \theta < 2\pi {\rm{\;and\;}}0 < r < 1\)

\(\begin{matrix} \int _{0}^{2\pi }\int _{0}^{1}2{{r}^{2}}\left( \frac{4\sqrt{3}}{3} \right)rdrd\theta \\ \left( \frac{8\sqrt{3}}{3} \right)\int _{0}^{2\pi }\left( \int _{0}^{1}{{r}^{3}}dr \right)d\theta \\ \left( \frac{8\sqrt{3}}{3} \right)\int _{0}^{2\pi }\left[ \frac{{{r}^{4}}}{4} \right]_{0}^{1}d\theta =\frac{4\sqrt{3}\pi }{3} \\ \end{matrix}\)

Hence, the required integral is \(\frac{{4\sqrt 3 \pi }}{3}\)