To solve the given double integral over the region enclosed by the ellipse, we can use a change of variables to transform the ellipse into a circle. This can be done by using the substitutions \( u = \frac{x}{3} \) and \( v = \frac{y}{6} \). The Jacobian determinant of this transformation will be used to adjust the integral accordingly.
We are given the double integral \(\iint_{R} 6x \, dA\), where \(R\) is the region enclosed by the ellipse \(\frac{x^2}{9} + \frac{y^2}{36} = 1\).
To simplify the integral, we use the substitutions \(u = \frac{x}{3}\) and \(v = \frac{y}{6}\). This transforms the ellipse into a unit circle \(u^2 + v^2 = 1\).
The Jacobian determinant for the transformation is:
\[
J = \begin{vmatrix}
\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\
\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}
\end{vmatrix} = \begin{vmatrix}
3 & 0 \\
0 & 6
\end{vmatrix} = 18
\]
Substituting \(x = 3u\) and \(y = 6v\) into the integrand \(6x\) and multiplying by the absolute value of the Jacobian determinant, we get:
\[
6x \cdot |J| = 6(3u) \cdot 18 = 324u
\]
To integrate over the unit circle, we convert to polar coordinates:
\[
u = r \cos \theta, \quad v = r \sin \theta
\]
The integrand in polar coordinates becomes:
\[
324u \cdot r = 324r^2 \cos \theta
\]
The double integral in polar coordinates is:
\[
\int_{0}^{2\pi} \int_{0}^{1} 324r^2 \cos \theta \, r \, dr \, d\theta
\]
Evaluating the integral:
\[
\int_{0}^{2\pi} \cos \theta \, d\theta = 0
\]
Thus, the entire integral evaluates to zero.
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