Question

Use Green's Theorem to prove the change of variables formula for a double integral (Formula 12.8.9) for the case where \(f(x,y) = 1\) :

Here \(R\) is the region in the \(xy\) -plane that corresponds to the region Sin the \(uv\)-plane under the transformation given by\(x = g\left( {u,{\rm{ }}v} \right),{\rm{ }}y = h\left( {u,{\rm{ }}v} \right)\)

(Hint: Note that the left side is \(A(R)\) and apply the first part of Equation 5. Convert the line integral over \(\partial R\) to a line integral over \(\partial S\) and apply Green's Theorem in the \(uv\)-plane.)

Short answer

Hence proved

\(\int_{R}{d}xdy=\iint_{S}{\left| \frac{\partial (x,y)}{\partial (u,v)} \right|}\)

Step-by-step solution

expression for Green's theorem

The expression for Green's theorem is\(A(R) = \int_{\partial R} x dy\)

Step 2:first part of Equation 5

The first part of Equation 5 is \(A(R) = \oint_C x dy\)

Therefore,

\(\int_R d xdy = \oint_{\partial R} x dy\)

Note that,

\(x = g(u,v);\quad y = h(u,v)\)

Therefore, by chain rule,

\(dy = \frac{{\partial h}}{{\partial u}}du + \frac{{\partial h}}{{\partial v}}dv\)

We have

\(\begin{array}{c}\oint_{\partial R} x dy = \oint_{\partial S} g (u,v)\left( {\frac{{\partial h}}{{\partial u}}du + \frac{{\partial h}}{{\partial v}}dv} \right)\\ = \oint_{\partial S} {\left( {g(u,v)\frac{{\partial h}}{{\partial u}}} \right)} du + \left( {g(u,v)\frac{{\partial h}}{{\partial v}}} \right)dv\end{array}\)

use green’s theorem

Use Green's theorem, we get

\(=\iint_{S}{\left[ \frac{\partial }{\partial u}\left( g(u,v)\frac{\partial h}{\partial v} \right)-\frac{\partial }{\partial v}\left( g(u,v)\frac{\partial h}{\partial u} \right) \right]}dA\)

Use chain rule

\(=\iint_{S}{\left[ \frac{\partial g}{\partial u}\cdot \frac{\partial h}{\partial v}+g\cdot \frac{\partial }{\partial u}\left( \frac{\partial h}{\partial v} \right) \right]}-\left[ \frac{\partial g}{\partial v}\cdot \frac{\partial h}{\partial u}+g\cdot \frac{\partial }{\partial v}\left( \frac{\partial h}{\partial u} \right) \right]dA\)

Rearrange

\(=\iint_{S}{\left[ \frac{\partial g}{\partial u}\cdot \frac{\partial h}{\partial v}-\frac{\partial g}{\partial v}\cdot \frac{\partial h}{\partial u} \right]}+g\left[ \frac{\partial }{\partial u}\left( \frac{\partial h}{\partial v} \right)-\frac{\partial }{\partial v}\left( \frac{\partial h}{\partial u} \right) \right]dA\)

Note that the blue part is zero

solve further

Therefore,

\(\begin{align} & \int_{R}{d}xdy=\iint_{S}{\left[ \frac{\partial g}{\partial u}\cdot \frac{\partial h}{\partial v}-\frac{\partial g}{\partial v}\cdot \frac{\partial h}{\partial u} \right]}dA \\ & =\iint_{S}{\left| \frac{\partial (g,h)}{\partial (u,v)} \right|} \end{align}\)

Note that \(x = g(u,v)\) and \(y = h(u,v)\)

Therefore,

\(\int_{R}{d}xdy=\iint_{S}{\left| \frac{\partial (x,y)}{\partial (u,v)} \right|}\)

Hence proved