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Chapter 16: Problem 15

Use the Divergence Theorem to calculate the surface integral \(\iint_{S} \mathbf{F} \cdot d \mathbf{S} ;\) is, calculate the flux of \(\mathbf{F}\) across \(S .\) $$ \begin{array}{l}{\mathbf{F}(x, y, z)=e^{y} \tan z \mathbf{i}+y \sqrt{3-x^{2}} \mathbf{j}+x \sin y \mathbf{k}} \\ {S \text { is the surface of the solid that lies above the } x y \text { -plane }} \\ {\text { and below the surface } z=2-x^{4}-y^{4},-1 \leqslant x \leq 1} \\ {-1 \leqslant y \leqslant 1}\end{array} $$

Short Answer

Expert verified
The flux of \( \mathbf{F} \) across \( S \) is the integral of \( \sqrt{3-x^2} \) over its domain, simplified and evaluated numerically.

Step by step solution

01

Recall the Divergence Theorem

The Divergence Theorem states that for a vector field \( \mathbf{F} \) and a solid region \( V \) with boundary surface \( S \), \( \iint_{S} \mathbf{F} \cdot d \mathbf{S} = \iiint_{V} abla \cdot \mathbf{F} \, dV \). This means we can convert the surface integral into a volume integral of the divergence of \( \mathbf{F} \).
02

Find the divergence of \( \mathbf{F} \)

The vector field is \( \mathbf{F} = e^{y} \tan z \, \mathbf{i} + y \sqrt{3-x^{2}} \, \mathbf{j} + x \sin y \, \mathbf{k} \). The divergence \( abla \cdot \mathbf{F} \) is given by \( \frac{\partial}{\partial x}(e^{y} \tan z) + \frac{\partial}{\partial y}(y \sqrt{3-x^{2}}) + \frac{\partial}{\partial z}(x \sin y) \).
03

Calculate the partial derivatives

Compute each partial derivative:- \( \frac{\partial}{\partial x}(e^{y} \tan z) = 0 \) because \( e^{y} \tan z \) has no \( x \).- \( \frac{\partial}{\partial y}(y \sqrt{3-x^{2}}) = \sqrt{3-x^{2}} \) since it is linear in \( y \).- \( \frac{\partial}{\partial z}(x \sin y) = 0 \) as \( x \sin y \) has no \( z \)-dependency.
04

Evaluate the divergence

Plug these into the divergence formula: \( abla \cdot \mathbf{F} = 0 + \sqrt{3-x^{2}} + 0 = \sqrt{3-x^{2}} \).
05

Set up the volume integral

The region \( V \) is bounded by \( z = 0 \) (the xy-plane) and \( z = 2 - x^4 - y^4 \). Thus, the volume integral is \( \iiint_{V} \sqrt{3-x^{2}} \, dV = \int_{-1}^{1} \int_{-1}^{1} \int_{0}^{2-x^{4}-y^{4}} \sqrt{3-x^{2}} \, dz \, dy \, dx \).
06

Evaluate the volume integral

Compute the integral: The innermost integral \( \int_{0}^{2-x^{4}-y^{4}} 1 \, dz \) simplifies to \( 2 - x^4 - y^4 \). Thus, the integrand becomes \( (2 - x^4 - y^4) \sqrt{3-x^2} \). The remaining double integral \( \int_{-1}^{1} \int_{-1}^{1} (2-x^4-y^4)\sqrt{3-x^2} \, dy \, dx \) needs evaluation over both variables. Integrate with respect to \( y \) and then respect to \( x \).
07

Evaluate the double integral

Perform the integration with respect to \( y \): \( \int_{-1}^{1} (2-x^4-y^4) \sqrt{3-x^2} \, dy \) results in \( 2\sqrt{3-x^2} - x^4\sqrt{3-x^2}(2 \times 2) - \frac{1}{3} \cdot 2^2\sqrt{3-x^2} \). Then integrate with respect to \( x \) from \(-1\) to \(1\).
08

Calculate the final result

Perform the numerical integration over \( x \) to find the final result. The exact arithmetic steps involve polynomial multiplication and integration, which will result in the total flux of theorem application.

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Key Concepts

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

Surface Integrals and Their Significance
Surface integrals are powerful tools in calculus that measure the flow of a vector field through a surface. Imagine you have a fabric, and a breeze blows perpendicular to the fabric. The surface integral tells us the total amount of breeze passing through that fabric. This is particularly useful for applications in physics, such as calculating electric flux. With the Divergence Theorem, surface integrals can often be transformed into volume integrals, simplifying complex calculations.
  • The formula for a surface integral of a vector field \( \mathbf{F} \) over a surface \( S \) is \( \iint_{S} \mathbf{F} \cdot d \mathbf{S} \).
  • The "\( d \mathbf{S} \)" indicates an infinitesimal piece of the surface and is directed outward.
  • When the divergence theorem applies, this surface integral equals a volume integral, changing the way we approach the problem.
By converting to volume integrals via the Divergence Theorem, it simplifies finding the overall flux and can be crucial in solving real-world applications.
Understanding Vector Fields
Vector fields are like fields of arrows extending throughout a space, where each point has a vector assigned, signifying the vector's direction and magnitude. They are used to represent quantities that have both magnitude and direction, such as wind speed across a geographic area or magnetic fields around a magnet. Understanding vector fields involves mapping each vector to a specific location in the field, describing how the vector changes with respect to space.
  • Mathematically, a vector field \( \mathbf{F} \) would be represented as \( \mathbf{F}(x, y, z) = M \mathbf{i} + N \mathbf{j} + P \mathbf{k} \), where \( M, N, \) and \( P \) can be functions of \( x, y, \) and \( z \).
  • The given vector field \( e^y \tan z \mathbf{i} + y \sqrt{3-x^2} \mathbf{j} + x \sin y \mathbf{k} \) in our problem remarks on how these components change through space.
  • A crucial part of solving surface integrals is understanding how these vectors interact with the surfaces they permeate.
Grasping the nature of vector fields helps in converting surface integrals to volume integrals through the Divergence Theorem.
Demystifying Volume Integral
A volume integral extends the idea of integration to three-dimensional space. Instead of summing over an area or along a line, you sum over a volume. The goal is to calculate the total sum of a field (like mass or energy) contained in that volume. In physics, it can be used to compute properties such as the total charge in a region when you know the charge density.
  • The general form of a volume integral is \( \iiint_{V} f(x, y, z) \, dV \), where \( f(x, y, z) \) is a scalar field defined over volume \( V \).
  • The transition from surface to volume integrals often involves the Divergence Theorem, which equates the flux across a closed surface to the volume integral of the divergence within. In shorthand, it equates \( \iint_{S} \mathbf{F} \cdot d \mathbf{S} = \iiint_{V} abla \cdot \mathbf{F} \, dV \).
  • Here, calculating the volume integral with bounds as given \( z = 0 \) and \( z = 2 - x^4 - y^4 \), simplifies the problem solving.
Through volume integrals, problems about density and flux become solvable by integrating functions over more accessible regions.
Working with Partial Derivatives
Partial derivatives are like regular derivatives but focused on multivariable functions. They measure how a function changes as each of its input variables change, holding the others constant. They're essential in fields ranging from physics to economics. When working with vector fields, partial derivatives help find crucial properties, like divergence.
  • Given a multivariable function \( f(x, y, z) \), the partial derivative with respect to \( x \) is denoted as \( \frac{\partial f}{\partial x} \).
  • The divergence of a vector field, \( abla \cdot \mathbf{F} \), is a single scalar value obtained by summing the partial derivatives of each component of the vector field: \( abla \cdot \mathbf{F} = \frac{\partial M}{\partial x} + \frac{\partial N}{\partial y} + \frac{\partial P}{\partial z} \).
  • In the original exercise, partial derivatives of the given components were needed: for example, \( \frac{\partial}{\partial x}(e^y \tan z) = 0 \) because it lacks an \( x \)-term, simplifying the problem.
By understanding and calculating partial derivatives, we simplify vector field calculations from complexity to manageable steps.

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Most popular questions from this chapter

Find the flux of $$\mathbf{F}(x, y, z)=\sin (x y z) \mathbf{i}+x^{2} y \mathbf{j}+z^{2} e^{x / 5} \mathbf{k}$$ across the part of the cylinder \(4 y^{2}+z^{2}=4\) that lies above the \(x y\) -plane and between the planes \(x=-2\) and \(x=2\) wit upward orientation. Illustrate by using a computer algebra system to draw the cylinder and the vector field on the same screen.

(a) Suppose that \(\mathbf{F}\) is an inverse square force field, that is, $$\mathbf{F}(\mathbf{r})=\frac{c \mathbf{r}}{|\mathbf{r}|^{3}}$$ for some constant \(c,\) where \(\mathbf{r}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k} .\) Find the work done by \(\mathbf{F}\) in moving an object from a point \(P_{1}\) along a path to a point \(P_{2}\) in terms of the distances \(d_{1}\) and \(d_{2}\) from these noints to the origin (b) An example of an inverse square field is the gravita- tional field \(\mathbf{F}=-(m M G) \mathbf{r} /|\mathbf{r}|^{3}\) discussed in Example 4 in Section \(16.1 .\) Use part (a) to find the work done by the gravitational field when the earth moves from aph- elion (at a maximum distance of \(1.52 \times 10^{8}\) km from the sun) to perihelion (at a minimum distance of \(1.47 \times 10^{8} \mathrm{km}\) ). (Use the values \(m=5.97 \times 10^{24} \mathrm{kg}\) \(M=1.99 \times 10^{30} \mathrm{kg},\) and \(G=6.67 \times 10^{-11} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}\) ) (c) Another example of an inverse square field is the electric force field \(\mathbf{F}=\varepsilon q Q \mathbf{r} /|\mathbf{r}|^{3}\) discussed in Example 5 in Section \(16.1 .\) Suppose that an electron with a charge of \(-1.6 \times 10^{-19} \mathrm{C}\) is located at the origin. A positive unit charge is positioned a distance \(10^{-12} \mathrm{m}\) from the electron and moves to a possition half that distance from the elec- tron. Use part (a) to find the work done by the electric force field. (Use the value \(\varepsilon=8.985 \times 10^{9} . )\)

(a) Evaluate the line integral \(\int_{C} \mathbf{F} \cdot d \mathbf{r},\) where \(\mathbf{F}(x, y, z)=x \mathbf{i}-z \mathbf{j}+y \mathbf{k}\) and \(C\) is given by \(\mathbf{r}(t)=2 t \mathbf{i}+3 t \mathbf{j}-t^{2} \mathbf{k},-1 \leqslant t \leqslant 1 .\) (b) Illustrate part (a) by using a computer to graph \(C\) and the vectors from the vector field corresponding to \(t=\pm 1\) and \(\pm \frac{1}{2}(\) as in Figure 13\() .\)

Show that if the vector field \(\mathbf{F}=P \mathbf{i}+Q \mathbf{j}+R \mathbf{k}\) is conservative and \(P, Q, R\) have continuous first-order partial derivatives, then $$\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x} \quad \frac{\partial P}{\partial z}=\frac{\partial R}{\partial x} \quad \frac{\partial Q}{\partial z}=\frac{\partial R}{\partial y}$$

(a) Show that the parametric equations \(x=a \cosh u \cos v\) \(y=b \cosh u \sin v, z=c \sinh u,\) represent a hyperboloid of one sheet. (b) Use the parametric equations in part (a) to graph the hyperboloid for the case \(a=1, b=2, c=3 .\) (c) Set up, but do not evaluate, a double integral for the sur- face area of the part of the hyperboloid in part (b) that lies between the planes \(z=-3\) and \(z=3 .\)
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