Questions: Find ∫∫ R 6x dA, where R is a region enclosed by the ellipse x^2/9 + y^2/36 = 1. (Give an exact answer. Use symbolic notation and fractions where needed.)

Solution Steps

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.

Final Answer