19. $iint_{r}x^{2}da$, where $r$ is the region bounded by the ellipse $9x^{2}+4y^{2}=36$; $x = 2u$, $y = 3v$

Explanation:

Step1: Find the Jacobian

First, find the partial - derivatives. Given $x = 2u$ and $y = 3v$, the Jacobian $J=\begin{vmatrix}\frac{\partial x}{\partial u}&\frac{\partial x}{\partial v}\\frac{\partial y}{\partial u}&\frac{\partial y}{\partial v}\end{vmatrix}=\begin{vmatrix}2&0\0&3\end{vmatrix}=6$.

Step2: Transform the ellipse equation

Substitute $x = 2u$ and $y = 3v$ into the ellipse equation $9x^{2}+4y^{2}=36$. We get $9(2u)^{2}+4(3v)^{2}=36$, which simplifies to $36u^{2}+36v^{2}=36$, or $u^{2}+v^{2}=1$.

Step3: Transform the integrand

Since $x = 2u$, then $x^{2}=(2u)^{2}=4u^{2}$. And $dA = |J|dudv=6dudv$.

Step4: Set up the double - integral in the new variables

The double - integral $\iint_{R}x^{2}dA$ becomes $\iint_{u^{2}+v^{2}\leq1}4u^{2}\cdot6dudv = 24\iint_{u^{2}+v^{2}\leq1}u^{2}dudv$.

Step5: Convert to polar coordinates

In polar coordinates, $u = r\cos\theta$, $v = r\sin\theta$, and $dudv = rdr d\theta$. The region $u^{2}+v^{2}\leq1$ becomes $r\leq1$ and $0\leq\theta\leq2\pi$. Also, $u^{2}=r^{2}\cos^{2}\theta$. So the integral is $24\int_{0}^{2\pi}\int_{0}^{1}(r^{2}\cos^{2}\theta)r drd\theta$.

Step6: Evaluate the integral

First, evaluate $\int_{0}^{1}r^{3}dr=\left[\frac{1}{4}r^{4}\right]{0}^{1}=\frac{1}{4}$. Then, evaluate $\int{0}^{2\pi}\cos^{2}\theta d\theta=\int_{0}^{2\pi}\frac{1 + \cos(2\theta)}{2}d\theta=\left[\frac{1}{2}\theta+\frac{1}{4}\sin(2\theta)\right]_{0}^{2\pi}=\pi$. The value of the integral is $24\times\frac{1}{4}\times\pi = 6\pi$.

Answer:

$6\pi$