Use a change of variables to evaluate the following integral.
ModifyingBelow Integral from nothing to nothing Integral from nothing to nothing Integral from nothing to nothing With Upper Dxy ​dV; D is bounded by the planes yminusxequals​0, yminusxequals4​, zminusyequals​0, zminusyequals1​, zequals​0, zequals7.
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Part 1
ModifyingB

To evaluate the given integral using a change of variables, we first need to understand the region \( D \) and then choose an appropriate transformation that simplifies the integration.

### Step 1: Identify the bounds of the region \( D \)

The region \( D \) is bounded by the following planes:
1. \( y - x = 0 \) (or \( y = x \))
2. \( y - x = 4 \) (or \( y = x + 4 \))
3. \( z - y = 0 \) (or \( z = y \))
4. \( z - y = 1 \) (or \( z = y + 1 \))
5. \( z = 0 \)
6. \( z = 7 \)

### Step 2: Choose a suitable change of variables

Given the equations of the planes, we can introduce new variables that align with these planes. Let's define:
\[
u = y - x
\]
\[
v = z - y
\]
\[
w = z
\]

This change of variables simplifies the boundaries as follows:
- \( u = 0 \) (corresponding to \( y = x \))
- \( u = 4 \) (corresponding to \( y = x + 4 \))
- \( v = 0 \) (corresponding to \( z = y \))
- \( v = 1 \) (corresponding to \( z = y + 1 \))
- \( w = 0 \) to \( w = 7 \) (corresponding to \( z = 0 \) to \( z = 7 \))

### Step 3: Express the volume element \( dV \) in terms of \( u, v, w \)

First, let's compute the Jacobian determinant of the transformation:
\[
\begin{aligned}
x &= x(u, v, w), \\
y &= y(u, v, w), \\
z &= z(u, v, w).
\end{aligned}
\]
From the substitution \( u = y - x \), \( v = z - y \), and \( w = z \), we solve:
\[
\begin{aligned}
y &= u + x, \\
z &= v + y = v + u + x.
\end{aligned}
\]
This implies:
\[
\begin{aligned}
x &= x, \\
y &= u + x, \\
z &= v + u + x.
\end{aligned}
\]

Taking partial derivatives to form the Jacobian matrix:
\[
\frac{\partial(x, y, z)}{\partial(u, v, w)} =
\begin{pmatrix}
1 & 0 & 0 \\
1 & 1 & 0 \\
1 & 1 & 1
\end{pmatrix}
\]
The determinant of this matrix is \( 1 \).

Thus, the volume element transforms as:
\[
dV = |1| \, du \, dv \, dw = du \, dv \, dw
\]

### Step 4: Set up the transformed integral

With the new variables, the integral becomes:
\[
\int_{w=0}^{7} \int_{v=0}^{1} \int_{u=0}^{4} \, du \, dv \, dw
\]

### Step 5: Evaluate the integral

Integrating step by step:
\[
\int_{w=0}^{7} \int_{v=0}^{1} \int_{u=0}^{4} du \, dv \, dw = \int_{w=0}^{7} \int_{v=0}^{1} [u]_{0}^{4} \, dv \, dw = \int_{w=0}^{7} \int_{v=0}^{1} 4 \, dv \, dw
\]
\[
= \int_{w=0}^{7} [4v]_{0}^{1} \, dw = \int_{w=0}^{7} 4 \, dw = 4w \bigg|_{0}^{7} = 4(7) = 28
\]

### Final Answer:
The value of the integral is \( \boxed{28} \).

Would you like further details or have any questions?

### Related Questions:
1. How does the Jacobian determinant affect the integral when using a change of variables?
2. Can the integral's value change if the transformation isn't one-to-one?
3. How do you determine the boundaries after a change of variables?
4. What are the applications of changing variables in double and triple integrals?
5. How would you handle a case where the Jacobian determinant is zero?
6. What are some examples of non-linear change of variables?
7. How do you interpret the geometric meaning of the volume element \( dV \)?
8. What happens if the region \( D \) was defined by different planes?

#### Tip: Always ensure the transformation is one-to-one within the region of integration when applying a change of variables.

Learn how to evaluate a triple integral using a change of variables technique. This problem involves multiple integrals in a bounded region defined by specific planes, applying the Jacobian determinant formula for transformation. Suitable for advanced undergraduate mathematics.

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