Find ∫∫D sin(y-x)/y+x dA; D: {y-x=0, y-x=1, y+x=2}

The problem requires finding the double integral \(\iint_D \frac{\sin(y-x)}{y+x} \, dA\), where \(D\) is the region bounded by the lines \(y-x=0\), \(y-x=1\), and \(y+x=2\).

### Step 1: Define the region \(D\)
Given the equations of the boundaries:

1. \(y - x = 0\) (i.e., \(y = x\))
2. \(y - x = 1\) (i.e., \(y = x + 1\))
3. \(y + x = 2\)

### Step 2: Change of Variables
We perform a change of variables to simplify the region and the integrand.

Let:
\[
u = y - x
\]
\[
v = y + x
\]

The Jacobian determinant for this transformation is:
\[
\frac{\partial (u, v)}{\partial (x, y)} = 
\begin{vmatrix}
-1 & 1 \\
1 & 1
\end{vmatrix} = (-1)(1) - (1)(1) = -2
\]
So, \( \frac{\partial (x, y)}{\partial (u, v)} = -\frac{1}{2} \).

The integrand \(\frac{\sin(y-x)}{y+x}\) becomes \(\frac{\sin(u)}{v}\) in the new variables.

### Step 3: Set up the new integration limits
The region \(D\) in the \((u,v)\) plane is bounded by:

- \(u = 0\) (from \(y-x=0\))
- \(u = 1\) (from \(y-x=1\))
- \(v = 2\) (from \(y+x=2\))

Given \(y = \frac{u+v}{2}\) and \(x = \frac{v-u}{2}\), the \(v\)-limits depend on \(u\):

- The line \(y+x=2\) (i.e., \(v = 2\)) bounds \(v\) above.

Since \(u = y - x\), for each fixed \(u\), \(v\) ranges from \(u\) to 2. Therefore, the bounds are:

\[
0 \leq u \leq 1
\]
\[
u \leq v \leq 2
\]

### Step 4: Evaluate the Integral
The double integral in the new variables is:

\[
\iint_D \frac{\sin(u)}{v} \left(-\frac{1}{2}\right) dv \, du
\]

This simplifies to:

\[
-\frac{1}{2} \int_{0}^{1} \int_{u}^{2} \frac{\sin(u)}{v} dv \, du
\]

### Step 5: Integrate with respect to \(v\)
The inner integral with respect to \(v\) is:

\[
\int_{u}^{2} \frac{1}{v} dv = \ln \left(\frac{v}{u}\right) \bigg|_{u}^{2} = \ln \left(\frac{2}{u}\right)
\]

Thus, the integral becomes:

\[
-\frac{1}{2} \int_{0}^{1} \sin(u) \ln\left(\frac{2}{u}\right) du
\]

### Step 6: Integrate with respect to \(u\)
This integral can be evaluated using integration by parts or numerical methods, but it is complex and does not simplify easily into elementary functions.

The integral is:

\[
-\frac{1}{2} \int_{0}^{1} \sin(u) \ln\left(\frac{2}{u}\right) du
\]

This is the final answer in terms of an integral. For most practical purposes, this would be left as is unless a specific numerical value is needed.

Would you like more details on how to evaluate the final integral or have any questions?

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### Related Questions:

1. How can we compute the integral \(\int_0^1 \sin(u) \ln\left(\frac{2}{u}\right) du\) using numerical methods?
2. What is the significance of using the Jacobian in coordinate transformations for double integrals?
3. How does the change of variables simplify the integration process in multivariable calculus?
4. Can the region \(D\) be described differently to simplify the integral evaluation?
5. How does integration by parts apply to integrals involving products of functions like \(\sin(u)\) and \(\ln(u)\)?

### Tip:
When dealing with complex regions in double integrals, consider using a change of variables to simplify both the region and the integrand.

Explore the step-by-step solution to the double integral ∫∫D sin(y-x)/y+x dA, where the region D is bounded by y-x=0, y-x=1, and y+x=2. Learn how to apply a change of variables and evaluate the integral using advanced calculus techniques. Suitable for undergraduate calculus students.

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