Math Problem Statement

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. Question content area bottom Part 1 ModifyingB

Solution

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

Step 1: Identify the bounds of the region DD

The region DD is bounded by the following planes:

  1. yx=0y - x = 0 (or y=xy = x)
  2. yx=4y - x = 4 (or y=x+4y = x + 4)
  3. zy=0z - y = 0 (or z=yz = y)
  4. zy=1z - y = 1 (or z=y+1z = y + 1)
  5. z=0z = 0
  6. z=7z = 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=yxu = y - x v=zyv = z - y w=zw = z

This change of variables simplifies the boundaries as follows:

  • u=0u = 0 (corresponding to y=xy = x)
  • u=4u = 4 (corresponding to y=x+4y = x + 4)
  • v=0v = 0 (corresponding to z=yz = y)
  • v=1v = 1 (corresponding to z=y+1z = y + 1)
  • w=0w = 0 to w=7w = 7 (corresponding to z=0z = 0 to z=7z = 7)

Step 3: Express the volume element dVdV in terms of u,v,wu, v, w

First, let's compute the Jacobian determinant of the transformation:

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.

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Math Problem Analysis

Mathematical Concepts

Multiple Integrals
Change of Variables
Jacobian Determinant

Formulas

Jacobian determinant formula

Theorems

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Suitable Grade Level

Advanced Undergraduate