https://file.aiquickdraw.com/mathbot/file/1729716132356cedyk0ox.jpg

Let's analyze the problem from the image step by step.

### 1. **Calculate the Jacobian**
The image defines the change of variables \( x = 2u \) and \( y = 3v \) to transform the coordinates of the region enclosed by the ellipse:

\[
\frac{x^2}{4} + \frac{y^2}{9} = 1
\]

The Jacobian determinant is given by:

\[
J(u, v) = \left| \frac{\partial(x, y)}{\partial(u, v)} \right|
\]

Where \( x = 2u \) and \( y = 3v \). Now let's compute the partial derivatives:

\[
\frac{\partial x}{\partial u} = 2, \quad \frac{\partial x}{\partial v} = 0
\]
\[
\frac{\partial y}{\partial u} = 0, \quad \frac{\partial y}{\partial v} = 3
\]

So the Jacobian determinant is:

\[
J(u, v) = \left| \begin{matrix} 2 & 0 \\ 0 & 3 \end{matrix} \right| = 2 \times 3 = 6
\]

Thus, \( J(u, v) = 6 \), which matches the provided answer.

---

### 2. **Rewrite the integral using the change of variables**
We are asked to rewrite the given integral over the region \( D \) (the upper part of the ellipse) by changing the variables to \( u \) and \( v \):

\[
\iint_D y \, dA = \iint_S f(u, v) \, du \, dv
\]

First, we need to express the function \( f(u, v) \) and determine the region \( S \).

#### a) **Function \( f(u, v) \):**

Since \( y = 3v \) and the Jacobian is 6, the new function is:

\[
f(u, v) = y \times \text{Jacobian} = 3v \times 6 = 18v
\]

Thus, \( f(u, v) = 18v \), which is already filled in the provided image.

#### b) **Region \( S \):**
The region \( D \) is the part of the ellipse \( \frac{x^2}{4} + \frac{y^2}{9} = 1 \) that lies above the \( x \)-axis.

Using the transformations \( x = 2u \) and \( y = 3v \), the equation of the ellipse becomes:

\[
\frac{(2u)^2}{4} + \frac{(3v)^2}{9} = 1 \quad \Rightarrow \quad u^2 + v^2 = 1
\]

Since we are only considering the part of the ellipse above the \( x \)-axis (i.e., \( y \geq 0 \)), we require \( v \geq 0 \).

Thus, the region \( S \) is described by \( u^2 + v^2 \leq 1 \) and \( v \geq 0 \). This corresponds to option **c**:

\[
S = \{(u, v) \mid u^2 + v^2 \leq 1, v \geq 0 \}
\]

---

### Summary:
1. The Jacobian is \( J(u, v) = 6 \).
2. The function \( f(u, v) = 18v \) and the region \( S \) is given by option **c**: \( \{(u, v) \mid u^2 + v^2 \leq 1, v \geq 0 \} \).

Would you like any further clarifications or more details on this? Here are some related questions:

1. How do you compute the Jacobian in more complex variable transformations?
2. What is the significance of the Jacobian in coordinate transformations?
3. Can you extend this method to non-elliptical regions?
4. How would the integral change if the region was the entire ellipse?
5. How does the orientation of the region affect the limits of integration?

**Tip**: When dealing with transformations, always visualize the transformed region to ensure correct limits for integration.

Let D be the region enclosed by the ellipse (x^2)/4 + (y^2)/9 = 1 above the x-axis. Calculate the Jacobian using the change of variables x = 2u and y = 3v. Rewrite the integral using the change of variables and determine f(u,v) and the region S.

Explore how to compute the Jacobian for a coordinate transformation and rewrite an integral using the change of variables for a region enclosed by an ellipse. This problem covers key concepts in multivariable calculus, including Jacobian determinants and integral transformations.

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