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In Exercises \(7-12,\) find a potential function \(f\) for the field \(\mathbf{F}\). $$\begin{aligned} \mathbf{F}=\frac{y}{1+x^{2} y^{2}} \mathbf{i}+\left(\frac{x}{1+x^{2} y^{2}}+\frac{z}{\sqrt{1-y^{2} z^{2}}}\right) \mathbf{j}+& \\ &\left(\frac{y}{\sqrt{1-y^{2} z^{2}}}+\frac{1}{z}\right) \mathbf{k} \end{aligned}$$

Short Answer

Expert verified
The potential function is \(f(x,y,z) = \frac{1}{y^2} \tan^{-1}(xy) - \sqrt{1 - y^2 z^2} + \arcsin(yz) + \ln|z| + K\).

Step by step solution

01

Identify the Components

The vector field can be split into three components: \[ F_1 = \frac{y}{1 + x^2 y^2}, \quad F_2 = \frac{x}{1 + x^2 y^2} + \frac{z}{\sqrt{1 - y^2 z^2}}, \quad F_3 = \frac{y}{\sqrt{1 - y^2 z^2}} + \frac{1}{z} \]
02

Integrate Respect to x

We start by integrating the first component \(F_1\) with respect to \(x\). \[ \int \frac{y}{1 + x^2 y^2} \, dx = \frac{1}{y^2} \tan^{-1}(xy) + C(y,z) \] where \(C(y,z)\) represents the integration constant whose variable dependencies are yet to be determined.
03

Integrate Respect to y

Next, integrate \(F_2\) with respect to \(y\), also adding any dependence from the previous step on \(C(y,z)\). \[ \int \left(\frac{x}{1 + x^2 y^2} + \frac{z}{\sqrt{1 - y^2 z^2}}\right) \, dy = \frac{1}{x} \tan^{-1}(xy) - \sqrt{1 - y^2 z^2} + g(x,z) \] where \(g(x,z)\) is the integration constant for this step.
04

Integrate Respect to z

Finally, integrate the third component \(F_3\) with respect to \(z\), accounting for any remaining dependencies from earlier calculations. \[ \int \left(\frac{y}{\sqrt{1 - y^2 z^2}} + \frac{1}{z}\right) \, dz = \arcsin(yz) + \ln|z| + h(x,y) \] where \(h(x,y)\) represents the integration constant with adjustments for dependencies.
05

Compare Results and Deduce Potential Function

Combine the results from each of the separate integrations to form the overall potential function \(f\). Ensure that all three terms are consistent and dependencies line up: \[ f(x,y,z) = \frac{1}{y^2} \tan^{-1}(xy) - \sqrt{1 - y^2 z^2} + \arcsin(yz) + \ln|z| + K \] where \(K\) is an overall constant of integration.

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Key Concepts

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

Vector Field
In mathematics, a vector field assigns a vector to each point in a subset of space. Imagine the wind, which, at each point in the atmosphere, has both direction and magnitude. A vector field is much like this: it provides a vector (a tool giving direction and magnitude) at every point in a space.

Our example vector field \( \mathbf{F} \) is a function of \( x, y, \) and \( z \) that attaches vectors to each point in \( \mathbb{R}^3 \). It's given by three components, each a function of the variables.
  • \( F_1 = \frac{y}{1 + x^2 y^2} \)
  • \( F_2 = \frac{x}{1 + x^2 y^2} + \frac{z}{\sqrt{1 - y^2 z^2}} \)
  • \( F_3 = \frac{y}{\sqrt{1 - y^2 z^2}} + \frac{1}{z} \)
Understanding vector fields is important in various fields like physics and engineering where they describe electromagnetic fields, fluid flows, and more. A potential function relates to these fields as it represents a scalar field whose gradient is the given vector field.
Integration
Integration is a core concept in calculus, often described as the process of finding the antiderivative. It's the reverse of differentiation and is used to determine the total size, area, volume, or other measures of scope.

In the context of finding a potential function for a vector field, integration helps us find a function whose gradient is the vector field. We integrate each component of the vector field with respect to its respective variable. Consider the first component of our field, \( F_1 = \frac{y}{1 + x^2 y^2} \). We integrate this with respect to \( x \): \[ \int \frac{y}{1 + x^2 y^2} \, dx = \frac{1}{y^2} \tan^{-1}(xy) + C(y,z) \] The result gives a piece of the potential function. The constants from integration, like \( C(y,z) \), may depend on other variables, influencing subsequent integration steps.
Arctangent Function
The arctangent function, \( \tan^{-1}(x) \), is the inverse operation of the tangent trigonometric function. While tangent takes an angle and gives you a ratio, arctangent takes a ratio and gives back the angle. It's valuable when working with integration, especially involving divisions of squares or products.

During the integration of \( F_1 \) in our vector field, we arrived at an arctangent expression:

\[ \int \frac{y}{1 + x^2 y^2} \, dx = \frac{1}{y^2} \tan^{-1}(xy) + C(y,z) \]

Here, the arctangent helps manage the complexity of the expression produced by the reciprocal and quadratic forms of \( x \) and other variables. It's a useful tool because it provides insights into the angles created within the components of the vectors in the field.
Inverse Sine Function
The inverse sine function, or arcsine, denoted as \( \arcsin(x) \), is the inverse of the sine function. Where sine gives us a ratio from an angle, arcsine returns the angle from the ratio. It's a crucial element in solving problems with trigonometric relationships, especially when you're reversing the sine operation to find angles.

In the calculation of our vector field's potential function, we employ the inverse sine function when integrating the third component \( F_3 \):

\[ \int \left(\frac{y}{\sqrt{1 - y^2 z^2}} + \frac{1}{z}\right) \, dz = \arcsin(yz) + \ln|z| + h(x,y) \]

The arcsine resolves expressions involved with the trigonometric identities and relationships. This function helps tackle expressions that include squares and square roots, often simplifying them into manageable angular components. It's especially useful for solving problems involving circular or oscillatory motion.

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Most popular questions from this chapter

a. A torus of revolution (doughnut) is obtained by rotating a circle \(C\) in the \(x z\) -plane about the \(z\) -axis in space. (See the accompanying figure.) If \(C\) has radius \(r>0\) and center \((R, 0,0),\) show that a parametrization of the torus is $$ \begin{aligned} \mathbf{r}(u, \boldsymbol{v})=&((R+r \cos u) \cos v) \mathbf{i} \\ &+((R+r \cos u) \sin v) \mathbf{j}+(r \sin u) \mathbf{k} \end{aligned} $$ where \(0 \leq u \leq 2 \pi\) and \(0 \leq v \leq 2 \pi\) are the angles in the figure. b. Show that the surface area of the torus is \(A=4 \pi^{2} R r\)

Find the flux of the field \(\mathbf{F}(x, y, z)=z^{2} \mathbf{i}+x \mathbf{j}-3 z \mathbf{k}\) outward through the surface cut from the parabolic cylinder \(z=4-y^{2}\) by the planes \(x=0, x=1,\) and \(z=0\) .

If \(\mathbf{F}=M \mathbf{i}+N \mathbf{j}+P \mathbf{k}\) is a differentiable vector field, we define the notation \(\mathbf{F} \cdot \nabla\) to mean $$M \frac{\partial}{\partial x}+N \frac{\partial}{\partial y}+P \frac{\partial}{\partial z}$$ For differentiable vector fields \(\mathbf{F}_{1}\) and \(\mathbf{F}_{2},\) verify the following identities. $$\begin{array}{l}{\text { a. } \nabla \times\left(\mathbf{F}_{1} \times \mathbf{F}_{2}\right)=\left(\mathbf{F}_{2} \cdot \nabla\right) \mathbf{F}_{1}-\left(\mathbf{F}_{1} \cdot \nabla\right) \mathbf{F}_{2}+\left(\nabla \cdot \mathbf{F}_{2}\right) \mathbf{F}_{1}-} \\\ {\quad\left(\nabla \cdot \mathbf{F}_{1}\right) \mathbf{F}_{2}} \\ {\mathbf{b} . \nabla\left(\mathbf{F}_{1} \cdot \mathbf{F}_{2}\right)=\left(\mathbf{F}_{1} \cdot \nabla\right) \mathbf{F}_{2}+\left(\mathbf{F}_{2} \cdot \nabla\right) \mathbf{F}_{1}+\mathbf{F}_{1} \times\left(\nabla \times \mathbf{F}_{2}\right)+} \\ {\mathbf{F}_{2} \times\left(\nabla \times \mathbf{F}_{1}\right)}\end{array}$$

Find the work done by \(\mathbf{F}\) in moving a particle once counterclockwise around the given curve. \(\mathbf{F}=2 x y^{3} \mathbf{i}+4 x^{2} y^{2} \mathbf{j}\) C: The boundary of the "triangular" region in the first quadrant enclosed by the \(x\) -axis, the line \(x=1,\) and the curve \(y=x^{3}\)

Verify Stokes' Theorem for the vector field \(\mathbf{F}=2 x y \mathbf{i}+x \mathbf{j}+\) \((y+z) \mathbf{k}\) and surface \(z=4-x^{2}-y^{2}, z \geq 0,\) oriented with unit normal n pointing upward.

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