Chapter 5: Problem 26
As needed, use a computer to plot graphs and to check values of integrals. Make the change of variables \(u=x-y, v=x+y,\) to evaluate the integral $$\int_{0}^{1} d y \int_{0}^{1-y} e^{(x-y) /(x+y)} d x.$$
Short Answer
Expert verified
\(\frac{e-1}{4}\)
Step by step solution
01
Change of Variables
Introduce the change of variables: let \( u = x - y \) and \( v = x + y \). This transforms the integral into a new coordinate system.
02
Find Jacobian Determinant
Calculate the Jacobian determinant \( J \). The transformation equations are \( x = \frac{u+v}{2} \) and \( y = \frac{v-u}{2} \). The Jacobian is calculated as follows:\[ J = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} = \begin{vmatrix} \frac{1}{2} & \frac{1}{2} \ -\frac{1}{2} & \frac{1}{2} \end{vmatrix} = \frac{1}{2}\end{vmatrix} + \frac{1}{2}\end{vmatrix} = \frac{1}{2} \]
03
Change Limits of Integration
The original integration region is defined by \( 0 \le y \le 1 \) and \( 0 \le x \le 1-y \). Under the transformation, these limits change to bottom-left being \( (u,v) = (0,0) \) and top-right being \( (u,v) = (1,1) \).
04
Transform the Integral
Transform the integral using the new variables and the Jacobian. The integral becomes: \[ \int_{0}^{1} d v \int_{0}^{v} e^{\frac{u}{v}} \left( \frac{1}{2} \right) d u\]
05
Evaluate the Inner Integral
Solve the inner integral: \[ \int_{0}^{v} e^{\frac{u}{v}} d u = v \left[ e^{\frac{u}{v}} \right]_{0}^{v} = v(e^{1} - 1)\]
06
Evaluate the Outer Integral
Now, evaluate the outer integral with the result from the inner integral: \[ \int_{0}^{1} v (e - 1) \left( \frac{1}{2} \right) d v = \frac{e-1}{2} \int_{0}^{1} v d v = \frac{e-1}{2} \left[\frac{v^2}{2} \right]_{0}^{1} = \frac{e-1}{4} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Jacobian Determinant
The Jacobian Determinant is a special function that helps us understand how areas or volumes transform under a change of variables. It's particularly useful in coordinate transformations of integrals.
When we switch from one set of variables to another, the Jacobian determinant tells us how an infinitesimal area changes. In simple terms, it shows the scale factor by which the area transforms.
To calculate the Jacobian determinant, we need partial derivatives of the new variables with respect to the old ones. For example, in our transformation equations, we have:
When we switch from one set of variables to another, the Jacobian determinant tells us how an infinitesimal area changes. In simple terms, it shows the scale factor by which the area transforms.
To calculate the Jacobian determinant, we need partial derivatives of the new variables with respect to the old ones. For example, in our transformation equations, we have:
- Let’s consider the transformations: \( x = \frac{u+v}{2} \) and \( y = \frac{v-u}{2} \).
- The Jacobian matrix \( J \) for this transformation looks like this: \[ J = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} = \begin{vmatrix} \frac{1}{2} & \frac{1}{2} \ -\frac{1}{2} & \frac{1}{2} \end{vmatrix} \]
- When you find the determinant of this matrix, you calculate: \[ \frac{1}{2}*\frac{1}{2} - (-\frac{1}{2}*\frac{1}{2}) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \]
- This tells us that the area in the new coordinate system is scaled by a factor of \frac{1}{2}.
Coordinate Transformation
Coordinate transformation is changing from one coordinate system to another. This technique simplifies complex integrals by making them easier to evaluate.
For instance, in our problem, we switched from \( (x, y) \) coordinates to \( (u, v) \) coordinates with the following transformations:
For instance, in our problem, we switched from \( (x, y) \) coordinates to \( (u, v) \) coordinates with the following transformations:
- \( u = x - y \)
- \( v = x + y \)
- Bottom-left corner: \( (u,v) = (0,0) \)
- Top-right corner: \( (u,v) = (1,1) \)
Double Integrals
Double integrals extend the concept of an integral to functions of two variables. They are used to compute volumes and areas.
We can then first solve the inner integral:\[ \int_{0}^{v} e^{\frac{u}{v}} d u = v \left[ e^{\frac{u}{v}} \right]_{0}^{v} = v(e^{1} - 1) \].
And finally solve the outer integral:\[ \int_{0}^{1} v(e - 1) \left( \frac{1}{2} \right) d v = \frac{e-1}{2} \int_{0}^{1} v d v = \frac{e-1}{2} \left[\frac{v^2}{2} \right]_{0}^{1} = \frac{e-1}{4} \].
This approach shows how coordinate transformations and Jacobian determinants simplify double integrals.
- When you set up a double integral, you integrate over a two-dimensional region. This involves an inner and outer integral.
- In our example, the original double integral was, \[ \int_{0}^{1} d y \int_{0}^{1-y} e^{(x-y) /(x+y)} d x \].
- After changing variables, this becomes two nested integrals: one inside the other. The limits are much simpler after transformation.
We can then first solve the inner integral:\[ \int_{0}^{v} e^{\frac{u}{v}} d u = v \left[ e^{\frac{u}{v}} \right]_{0}^{v} = v(e^{1} - 1) \].
And finally solve the outer integral:\[ \int_{0}^{1} v(e - 1) \left( \frac{1}{2} \right) d v = \frac{e-1}{2} \int_{0}^{1} v d v = \frac{e-1}{2} \left[\frac{v^2}{2} \right]_{0}^{1} = \frac{e-1}{4} \].
This approach shows how coordinate transformations and Jacobian determinants simplify double integrals.