Questions: This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. Tutorial Exercise Evaluate the given integral by making an appropriate change of variables. ∬ over R (x-2y)/(3x-y) dA, where R is the parallelogram enclosed by the lines x-2y=0, x-2y=2, 3x-y=1, and 3x-y=2 Step 1 Let u=x-2y and v=3x-y. Then in terms of u and v, x=(2/5)v- square and y=(1/5)v- square . Therefore, ∂(x, y)/∂(u, v) = (∂x/∂u) (∂x/∂v) (∂y/∂u) (∂y/∂v) = (-1/5) (2/5) (-3/5) (1/5) = square

This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. Tutorial Exercise Evaluate the given integral by making an appropriate change of variables. ∬ over R (x-2y)/(3x-y) dA, where R is the parallelogram enclosed by the lines x-2y=0, x-2y=2, 3x-y=1, and 3x-y=2 Step 1 Let u=x-2y and v=3x-y. Then in terms of u and v, x=(2/5)v- square and y=(1/5)v- square . Therefore, ∂(x, y)/∂(u, v) = (∂x/∂u) (∂x/∂v) (∂y/∂u) (∂y/∂v) = (-1/5) (2/5) (-3/5) (1/5) = square
Transcript text: 5. [-/1 Points] DETAILS MY NOTES SCALCET9 15.9.025.MI.SA. This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. Tutorial Exercise Evaluate the given integral by making an appropriate change of variables. $\iint_{R} \frac{x-2 y}{3 x-y} d A$, where $R$ is the parallelogram enclosed by the lines $x-2 y=0, x-2 y=2,3 x-y=1$, and $3 x-y=2$ Step 1 Let $u=x-2 y$ and $v=3 x-y$. Then in terms of $u$ and $v, x=\frac{2}{5} v-$ $\square$ and $y=\frac{1}{5} v-$ $\square$ . Therefore, \[ \frac{\partial(x, y)}{\partial(u, v)}=\left|\begin{array}{ll} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{array}\right|=\left|\begin{array}{cc} -\frac{1}{5} & \frac{2}{5} \\ -\frac{3}{5} & \frac{1}{5} \end{array}\right|=\square \] Submit Skip (you cannot come back)
failed

Solution

failed
failed

Solution Steps

To solve the given integral, we need to perform a change of variables using the provided transformations \( u = x - 2y \) and \( v = 3x - y \). First, express \( x \) and \( y \) in terms of \( u \) and \( v \). Then, compute the Jacobian determinant of the transformation to adjust the differential area element \( dA \). Finally, substitute these into the integral and evaluate it over the new region in the \( uv \)-plane.

Step 1: Change of Variables

We start with the transformations defined by: \[ u = x - 2y \] \[ v = 3x - y \] Solving these equations for \( x \) and \( y \) gives: \[ x = -\frac{u}{5} + \frac{2v}{5} \] \[ y = -\frac{3u}{5} + \frac{v}{5} \]

Step 2: Jacobian Determinant

Next, we compute the Jacobian matrix of the transformation: \[ \frac{\partial(x, y)}{\partial(u, v)} = \begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{bmatrix} = \begin{bmatrix} -\frac{1}{5} & \frac{2}{5} \\ -\frac{3}{5} & \frac{1}{5} \end{bmatrix} \] Calculating the determinant of this matrix yields: \[ \text{det} = \left(-\frac{1}{5}\right) \left(\frac{1}{5}\right) - \left(-\frac{3}{5}\right) \left(\frac{2}{5}\right) = \frac{1}{5} \]

Final Answer

The expressions for \( x \) and \( y \) in terms of \( u \) and \( v \) are: \[ x = -\frac{u}{5} + \frac{2v}{5}, \quad y = -\frac{3u}{5} + \frac{v}{5} \] The Jacobian determinant is: \[ \frac{\partial(x, y)}{\partial(u, v)} = \frac{1}{5} \] Thus, the final boxed answers are: \[ \boxed{x = -\frac{u}{5} + \frac{2v}{5}} \quad \text{and} \quad \boxed{y = -\frac{3u}{5} + \frac{v}{5}} \quad \text{and} \quad \boxed{\text{det} = \frac{1}{5}} \]

Was this solution helpful?
failed
Unhelpful
failed
Helpful

Questions asked by other users

failedAnswered
Use a right triangle to write the following expression as an algebraic expression. Assume that x is positive and that the given inverse trigonometric function is defined for the expression in x. sec(cos^(-1) (2/x))
failedAnswered
Question 3 1 pts ATSI report of A / H2 S indicates an organism that is positive for: Glucose fermentation Lactose/sucrose utilization H2 S production Both lactose/sucrose utilization and H2 S production Both glucose fermentation and H2 S production
failedAnswered
Step 2. Graphing the data. The graph of the data that gives a straight line indicates the correct integrated rate law. Data from a single experiment are graphed three ways. The graph of [A] vs t is curve sloping downward. The graph of ln [A] vs t is a line with a negative slope, and the graph of 1 /[A] vs t is a curve sloping upward. Based on these descriptions, what rate law applies to this reaction? A The reaction is first order and the rate law is Rate =k[A] B The reaction is second order and the rate law is Rate =k[A]^2 C The reaction is zeroth order and the rate law is Rate =k
failedAnswered
A block with mass m1 hangs from a rope that is extended over an ideal pulley and attached to a second block with mass m2 that sits on a ledge slanted at an angle of 20° (see figure below). Suppose the system of blocks is initially held motionless and, when released, begins to accelerate. (a) If m1=7.25 kg, m2=3.00 kg, and the magnitude of the acceleration of the blocks is 0.170 m / s^2, find the magnitude of the kinetic frictional force between the second block and the ledge. (b) What is the value of the coefficient of kinetic friction between the block and the ledge?
failedAnswered
Unit Exam - Populations r=(CBR+IMA)-(CDR+EM) CBR=1 % IMA=1 % CDR=1 % EM=3 % What is r? -2 % 4 % 6 %