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Chapter 19: Problem 11

Compute the flux of \(\vec{F}\) through the surface \(S,\) which is the part of the graph of \(z=f(x, y)\) corresponding to region \(R,\) oriented upward. $$\begin{array}{l}\vec{F}(x, y, z)=\cos y \vec{i}+z \vec{j}+\vec{k} \\\\\quad f(x, y)=x^{2}+2 y \\\\\quad R: 0 \leq x \leq 1,0 \leq y \leq 1\end{array}$$

Short Answer

Expert verified
The flux through the surface is approximately -1.174.

Step by step solution

01

Understanding the Problem

We need to compute the flux of the vector field \( \vec{F} = \cos y \vec{i} + z \vec{j} + \vec{k} \) through the surface defined by \( z = f(x, y) = x^2 + 2y \) over the region \( R \) where \( 0 \leq x \leq 1 \) and \( 0 \leq y \leq 1 \) with the surface oriented upwards.
02

Surface Parameterization

The surface \( S \) can be parameterized by considering \( (x, y) \) as parameters. Thus: \( \vec{r}(x, y) = \langle x, y, x^2 + 2y \rangle \).
03

Computing the Normal Vector

The normal vector is obtained by computing the cross product of the partial derivatives of \( \vec{r} \) with respect to \( x \) and \( y \). Let \( \vec{r}_x = \langle 1, 0, 2x \rangle \) and \( \vec{r}_y = \langle 0, 1, 2 \rangle \). Compute \( \vec{n} = \vec{r}_x \times \vec{r}_y = \langle -2x, -2, 1 \rangle \).
04

Dot Product with \( \vec{F} \)

Evaluate \( \vec{F} \cdot \vec{n} \). We have \( \vec{F}(x, y, f(x, y)) = \cos y \vec{i} + (x^2 + 2y) \vec{j} + \vec{k} \). Thus, \( \vec{F} \cdot \vec{n} = (\cos y)(-2x) + (x^2 + 2y)(-2) + (1) \). Simplify this to get: \(-2x \cos y - 2x^2 - 4y + 1 \).
05

Set Up the Double Integral

Compute the integral of \( \vec{F} \cdot \vec{n} \) over region \( R \). The flux \( \Phi \) is given by:\[ \Phi = \int_0^1 \int_0^1 (-2x \cos y - 2x^2 - 4y + 1) \,dy \, dx \]
06

Evaluate the Inner Integral

Integrate with respect to \( y \):\[ \int_0^1 (-2x \cos y - 4y + 1) \, dy = \left[-2x \sin y - 2y^2 + y \right]_0^1 = -2x \sin 1 - 2 + 1 \]
07

Evaluate the Outer Integral

Substitute the result of the inner integral and evaluate:\[ \Phi = \int_0^1 (-2x \sin 1 - 2 + 1 - 2x^2) \, dx \]Simplify and integrate:\[ = \left[-x^2 \sin 1 - x + \frac{2x^3}{3}\right]_0^1 = -\sin 1 - 1 + \frac{2}{3} \]
08

Compute the Flux

Calculate the value using estimates:\[ \Phi = -\sin 1 - 1 + \frac{2}{3} \]Approximate \(-\sin 1 \sim -0.841\) to find:\[ \Phi \approx -0.841 - 1 + 0.667 = -1.174 \]

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Key Concepts

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

Vector Field
A vector field is a function that assigns a vector to every point in space. In this exercise, the vector field is defined as \( \vec{F}(x, y, z) = \cos y \vec{i} + z \vec{j} + \vec{k} \). This means that at any point \((x, y, z)\) in space, the vector field produces a vector with a component \(\cos y\) in the \(x\)-direction, \(z\) in the \(y\)-direction, and a constant 1 in the \(z\)-direction. Understanding this concept is crucial because the flux calculation depends on how these vectors flow through a given surface.
  • A vector field can represent physical phenomena such as velocity fields in fluids or the magnetic field in space.
  • The component functions (\(\cos y, z, 1\)) determine the vectors' directions and magnitudes at each point.
Visualizing a vector field helps us understand how different vectors interact with a surface for calculating flux.
Partial Derivatives
Partial derivatives involve differentiating a function concerning one of its variables while holding the others constant. In this problem, the function \( f(x, y) = x^2 + 2y \) is differentiated to help in parameterizing the surface and finding the normal vector.
  • The partial derivative of \( f(x, y) \) with respect to \(x\), represented as \( f_x(x, y) \), is \( 2x \).
  • The partial derivative with respect to \(y\), represented as \( f_y(x, y) \), is 2.
These derivatives are vital as they are used to compute the cross product required for the normal vector to the surface. Calculating partial derivatives allows us to understand changes in the surface relating to each variable individually, essential for surface parameterization.
Surface Parameterization
Surface parameterization is a way to represent a surface using a set of parameters, usually two, for a surface in three dimensions. Here, the surface \( S \) is parameterized using the coordinates \( (x, y) \). This leads to a vector function \( \vec{r}(x, y) = \langle x, y, x^2 + 2y \rangle \), which maps 2D points to 3D space.
  • This parameterization is used to express the surface \( z = f(x, y) = x^2 + 2y \) in terms of \( x \) and \( y \).
  • It helps in the computation of the normal vector essential for flux calculations.
By parameterizing the surface, we transform the problem into a form where we can apply calculus to find physical quantities like flux.
Double Integral
The double integral is a way to sum up values over a 2-dimensional region. In this exercise, it's used to compute the total flux through the surface by integrating the expression \( \vec{F} \cdot \vec{n} \) over the region \(R\). The formula for flux \( \Phi \) is:\[\Phi = \int_0^1 \int_0^1 (-2x \cos y - 2x^2 - 4y + 1) \,dy \, dx\]
  • This requires evaluating two separate integrals, one nested inside the other, known as iterated integrals.
  • The inner integral sums up contributions over \(y\) for a fixed \(x\), while the outer integral aggregates these over \(x\).
Double integrals are vital for calculating area, volume, and in this case, the flux through a surface. They allow us to accumulate the value from every part of the defined region.

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Most popular questions from this chapter

Compute the flux of \(\vec{F}\) through the spherical surface, \(S\). \(\vec{F}=z \vec{k}\) and \(S\) is the upper hemisphere of radius 2 centered at the origin, oriented outward.

Compute the flux of the vector field \(\vec{F}\) through the surface \(S\). \(\vec{F}=x \vec{i}+y \vec{j}+z \vec{k}\) and \(S\) is a closed cylinder of radius 2 centered on the \(y\) -axis, with \(-3 \leq y \leq 3,\) and oriented outward.

Compute the flux of \(\vec{F}\) through the spherical surface, \(S\). \(\vec{F}=y \vec{i}-x \vec{j}+z \vec{k}\) and \(S\) is the spherical cap given by \(x^{2}+y^{2}+z^{2}=1, z \geq 0,\) oriented upward.

Are the statements true or false? Give reasons for your answer. If \(S\) is the part of the graph of \(z=f(x, y)\) above \(a \leq x \leq b, c \leq y \leq d,\) then \(S\) has surface area \(\int_{a}^{b} \int_{c}^{d}\sqrt{\left(f_{x}\right)^{2}+\left(f_{y}\right)^{2}+1} d x d y\)

Explain what is wrong with the statement. $$\begin{aligned} &\text { For } \vec{F}(x, y, z)=\left(x^{2}+y\right) \vec{i}+(2 y+z) \vec{j}-z^{2} \vec{k} \text { we have }\\\ &\operatorname{div} \vec{F}=2 x \vec{i}+2 \vec{j}-2 z \vec{k} \end{aligned}$$
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