Answer :
Answer:
Definition:
The Jacobian of the transformation x = f(r, θ) and y = g(r ,θ) is:
[tex]\frac{\partial (x,y) }{\partial (r,\theta)}=\begin{vmatrix}\frac{\partial x }{\partial r} & \frac{\partial x }{\partial \theta} \\ \frac{\partial y }{\partial r} & \frac{\partial y }{\partial \theta}\end{vmatrix}[/tex] =[tex]\frac{\partial x}{\partial r}\cdot \frac{\partial y}{\partial \theta}-\frac{\partial x}{\partial \theta}\frac{\partial y}{\partial r}[/tex] ......[1]
Given: [tex]x=6e^{-3r}\sin 2\theta[/tex] and [tex]y=e^{3r}\cos 2\theta[/tex]
then,
[tex]\frac{\partial x}{\partial r} = -18e^{-3r}\sin 2\theta[/tex]
[tex]\frac{\partial x}{\partial \theta} = 6e^{-3r}(2 \cos 2\theta) = 12e^{-3r}\cos 2\theta[/tex]
[tex]\frac{\partial y}{\partial r} = 3e^{3r}\cos 2\theta[/tex]
and
[tex]\frac{\partial y}{\partial \theta} = e^{3r}(-2 \sin 2\theta) = -2e^{3r}\sin 2\theta[/tex]
Substitute these value in [1] ;
[tex]\begin{vmatrix}\frac{\partial x }{\partial r} & \frac{\partial x }{\partial \theta} \\ \frac{\partial y }{\partial r} & \frac{\partial y }{\partial \theta}\end{vmatrix}=\begin{vmatrix}-18e^{-3r}\sin 2\theta & 12e^{-3r}\cos 2\theta\\ 3e^{3r}\cos 2\theta &- 2e^{3r}\sin 2\theta\end{vmatrix}[/tex]
=[tex](-18e^{-3r}\sin 2\theta)\cdot(-2e^{3r}\sin 2\theta)-(12e^{-3r}\cos 2\theta)\cdot(3e^{3r}\cos 2\theta)[/tex]
=[tex](-18 \cdot -2)e^{-3r+3r} \sin^2 2\theta - (12 \cdot 3)e^{-3r+3r} \cos^2 2\theta[/tex]
On simplify:
[tex]\begin{vmatrix}-18e^{-3r}\sin 2\theta & 12e^{-3r}\cos 2\theta\\ 3e^{3r}\cos 2\theta &- 2e^{3r}\sin 2\theta\end{vmatrix} = 36 \sin^2 2\theta -36 \cos^2 2\theta=36(\sin^2 2\theta-\cos^2 2\theta)[/tex]
=[tex]-36 (\cos 4\theta)[/tex]
[ Use [tex]\cos^2 2\theta -\sin^2 2\theta = \cos(4\theta)[/tex]]
therefore, the jacobian transformation of [tex]x=6e^{-3r}\sin 2\theta[/tex] and [tex]y=e^{3r}\cos 2\theta[/tex] is; [tex]-36 (\cos 4\theta)[/tex]