Math Problem Statement

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.

Solution

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

1. Calculate the Jacobian

The image defines the change of variables x=2ux = 2u and y=3vy = 3v to transform the coordinates of the region enclosed by the ellipse:

x24+y29=1\frac{x^2}{4} + \frac{y^2}{9} = 1

The Jacobian determinant is given by:

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

Where x=2ux = 2u and y=3vy = 3v. Now let's compute the partial derivatives:

xu=2,xv=0\frac{\partial x}{\partial u} = 2, \quad \frac{\partial x}{\partial v} = 0 yu=0,yv=3\frac{\partial y}{\partial u} = 0, \quad \frac{\partial y}{\partial v} = 3

So the Jacobian determinant is:

J(u,v)=2003=2×3=6J(u, v) = \left| \begin{matrix} 2 & 0 \\ 0 & 3 \end{matrix} \right| = 2 \times 3 = 6

Thus, J(u,v)=6J(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 DD (the upper part of the ellipse) by changing the variables to uu and vv:

DydA=Sf(u,v)dudv\iint_D y \, dA = \iint_S f(u, v) \, du \, dv

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

a) Function f(u,v)f(u, v):

Since y=3vy = 3v and the Jacobian is 6, the new function is:

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

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

b) Region SS:

The region DD is the part of the ellipse x24+y29=1\frac{x^2}{4} + \frac{y^2}{9} = 1 that lies above the xx-axis.

Using the transformations x=2ux = 2u and y=3vy = 3v, the equation of the ellipse becomes:

(2u)24+(3v)29=1u2+v2=1\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 xx-axis (i.e., y0y \geq 0), we require v0v \geq 0.

Thus, the region SS is described by u2+v21u^2 + v^2 \leq 1 and v0v \geq 0. This corresponds to option c:

S={(u,v)u2+v21,v0}S = \{(u, v) \mid u^2 + v^2 \leq 1, v \geq 0 \}


Summary:

  1. The Jacobian is J(u,v)=6J(u, v) = 6.
  2. The function f(u,v)=18vf(u, v) = 18v and the region SS is given by option c: {(u,v)u2+v21,v0}\{(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.

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Math Problem Analysis

Mathematical Concepts

Elliptic Regions
Coordinate Transformation
Jacobian Determinant
Multivariable Calculus
Integral Transformation

Formulas

Jacobian: J(u,v) = |∂(x,y)/∂(u,v)|
Ellipse equation: (x^2)/4 + (y^2)/9 = 1
Variable transformations: x = 2u, y = 3v
Integral transformation: ∫∫_D y dA = ∫∫_S f(u,v) du dv

Theorems

Jacobian determinant for coordinate transformation
Area transformation in multivariable calculus

Suitable Grade Level

Undergraduate Calculus