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Chapter 15: Problem 14

Show that the differential forms in the integrals are exact. Then evaluate the integrals. $$\int_{(1,1,2)}^{(3,5,0)} y z d x+x z d y+x y d z$$

Short Answer

Expert verified
The integral evaluates to -2.

Step by step solution

01

Identifying the Differential Form

Consider the differential form given by \( \omega = yz \, dx + xz \, dy + xy \, dz \).We need to determine whether this form is exact.
02

Check for Exactness Using Clairaut's Theorem

A differential form \(P(x,y,z)\ dx + Q(x,y,z)\ dy + R(x,y,z)\ dz\) is exact if there exists a function \(f\) such that \[ \frac{\partial f}{\partial x} = P, \quad \frac{\partial f}{\partial y} = Q, \quad \frac{\partial f}{\partial z} = R \] and if the mixed partial derivatives satisfy:\( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \),\( \frac{\partial P}{\partial z} = \frac{\partial R}{\partial x} \) and\( \frac{\partial Q}{\partial z} = \frac{\partial R}{\partial y} \).Calculate each: - \( \frac{\partial P}{\partial y} = z \), - \( \frac{\partial Q}{\partial x} = z \),- \( \frac{\partial P}{\partial z} = y \), - \( \frac{\partial R}{\partial x} = y \),- \( \frac{\partial Q}{\partial z} = x \), - \( \frac{\partial R}{\partial y} = x \).All are equal, indicating \(\omega\) is exact.
03

Find Potential Function

Since \(\omega\) is an exact form, there exists a function \(f(x, y, z)\) from which \(\omega\) is derived. Find \(f\) such that:\[ \frac{\partial f}{\partial x} = yz, \quad \frac{\partial f}{\partial y} = xz, \quad \frac{\partial f}{\partial z} = xy \]Integrate \(\frac{\partial f}{\partial x} = yz\) with respect to \(x\):\[ f(x, y, z) = xyz + g(y, z) \]Now, use \(\frac{\partial f}{\partial y} = xz\):\[ \frac{\partial}{\partial y}(xyz + g(y, z)) = xz \implies xz + \frac{\partial g}{\partial y} = xz \]Thus, \(\frac{\partial g}{\partial y} = 0\) implies \( g(y, z) = h(z) \).Finally, use \(\frac{\partial f}{\partial z} = xy\):\[ \frac{\partial}{\partial z}(xyz + h(z)) = xy \implies xy + h'(z) = xy \]Thus, \(h'(z) = 0\), indicating \(h(z)\) is constant.Therefore, \(f(x, y, z) = xyz + C\).
04

Evaluate the Integral Using the Potential Function

Now that we have the potential function \(f(x, y, z) = xyz + C\), evaluate the line integral using the fundamental theorem for line integrals:\[ \int_{(1,1,2)}^{(3,5,0)} \omega = f(3,5,0) - f(1,1,2) \]Calculate:\[ f(3,5,0) = 3 \times 5 \times 0 + C = 0 + C \]\[ f(1,1,2) = 1 \times 1 \times 2 + C = 2 + C \]Therefore, the integral evaluates to:\[ (0 + C) - (2 + C) = -2 \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Line Integrals
A line integral extends the concept of integrating a function to integration along a path or curve. Instead of integrating over a simple interval, a line integral considers a path in a multi-dimensional space.
When you perform a line integral, you are summing up values along a curve or path, often with each point contributing differently depending on its position on the path.
This makes line integrals especially useful for finding the work done by a force field, or the flow of a fluid along a path.
  • A line integral of a function, say, along a curve in a vector field, is denoted by \( \int_C \mathbf{F} \cdot d\mathbf{r} \).
  • This involves the dot product of the vector field \( \mathbf{F} \) and an infinitesimal vector \( d\mathbf{r} \) along the curve \( C \).
  • For exact forms, line integrals simplify, because they can be evaluated using a potential function.
In the exercise above, the differential form representing the line integral is found to be exact, allowing us to use a potential function to evaluate the integral easily.
Potential Functions
A potential function is a scalar function from which a given vector field can be derived as a gradient. If a vector field \( \mathbf{F} \) is the gradient of a scalar function \( f \), i.e., \( \mathbf{F} = abla f \), the function \( f \) is called a potential function.
This is extremely useful because when a differential form is exact (and hence a gradient field), calculating integrals becomes straightforward using potential functions.
  • If \( \omega = P\,dx + Q\,dy + R\,dz \) is an exact differential form, then there exists a potential function \( f(x,y,z) \) such that \( \omega = df = abla f \cdot d\mathbf{r} \).
  • In the problem, after determining that the given differential form is exact, we derived the potential function \( f(x, y, z) = xyz + C \).
  • This makes evaluating the line integral as simple as finding the difference \( f(b) - f(a) \) at the boundaries of the path.
In this scenario, potential functions allow us to convert the challenge of integrating over a path into a simpler subtraction problem using the values of \( f \).
Clairaut's Theorem
Clairaut's theorem, often used in the context of exact differential equations, provides a convenient criterion for checking the exactness of differential forms. This theorem states that a differential form \( P(x,y)\,dx + Q(x,y)\,dy \) is exact if the mixed partial derivatives are equal.
For functions involving three variables, the theorem extends as:
\( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \)
\( \frac{\partial P}{\partial z} = \frac{\partial R}{\partial x} \)
\( \frac{\partial Q}{\partial z} = \frac{\partial R}{\partial y} \)
  • It is essential for verifying whether a given differential form can be considered exact.
  • With exact forms, there is an associated potential function, simplifying many integral computations.
  • In the exercise, Clairaut's theorem was used to ensure that all mixed partial derivatives were equal, confirming the exactness of the form.
This allowed us to proceed confidently in finding the potential function and evaluating the line integral.

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Most popular questions from this chapter

Calculating Area with Green's Theorem If a simple closed curve C in the plane and the region \(R\) it encloses satisfy the hypotheses of Green's Theorem, the area of \(R\) is given by Green's Theorem Area Formula Area of \(R=\frac{1}{2} \oint_{c} x d y-y d x\) The reason is that, by Equation (4) run backward, Area of \(R=\iint_{k} d y d x=\iint_{k}\left(\frac{1}{2}+\frac{1}{2}\right) d y d x\) $$=\oint_{c} \frac{1}{2} x d y-\frac{1}{2} y d x$$ Use the Green's Theorem area formula given above to find the areas of the regions enclosed by the curves in Exercises \(31-34\) The ellipse \(\mathbf{r}(t)=(a \cos t) \mathbf{i}+(b \sin t) \mathbf{j}, \quad 0 \leq t \leq 2 \pi\)

Find a vector field with twice-differentiable components whose curl is \(x \mathbf{i}+y \mathbf{j}+z \mathbf{k},\) or prove that no such field exists.

Use the Divergence Theorem to find the outward flux of \(\mathbf{F}\) across the boundary of the region \(D\). Portion of sphere \(\quad \mathbf{F}=x^{2} \mathbf{i}-2 x y \mathbf{j}+3 x z \mathbf{k}\) D: The region cut from the first octant by the sphere \(x^{2}+y^{2}+\) \(z^{2}=4\)

Show that if \(R\) is a region in the plane bounded by a piecewise smooth, simple closed curve \(C\), then $$ \text { Area of } R=\oint x d y=-\oint y d x $$

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