/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 Use a parametrization to find th... [FREE SOLUTION] | 91Ó°ÊÓ

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Use a parametrization to find the flux \(\iint_{S} \mathbf{F} \cdot \mathbf{n} d \sigma\) across the surface in the specified direction. Parabolic cylinder \(\quad \mathbf{F}=z^{2} \mathbf{i}+x \mathbf{j}-3 z \mathbf{k} \quad\) outward \(\quad\) (normal away from the \(x\) -axis) through the surface cut from the parabolic cylinder \(z=4-y^{2}\) by the planes \(x=0, x=1,\) and \(z=0\).

Short Answer

Expert verified
The flux through the surface is -32.

Step by step solution

01

Parametrize the Surface

The surface is given by the parabolic cylinder equation \(z = 4 - y^2\). We need to parametrize the regions defined by the constraints \(x = 0\), \(x = 1\), and \(z = 0\). The surface can be parametrized by two parameters \(y\) and \(x\) as follows: \(\mathbf{r}(x, y) = (x, y, 4-y^2)\). \((x, y)\) varies over the region where \(0 \leq x \leq 1\) and \(-2 \leq y \leq 2\).
02

Compute the Normal Vector

To find the outward flux, we need the normal vector. We first compute the partial derivatives of \(\mathbf{r}\):\[ \mathbf{r}_x = (1, 0, 0) \] and\[ \mathbf{r}_y = (0, 1, -2y). \]The normal vector is given by the cross product of these two partial derivatives: \[ \mathbf{n} = \mathbf{r}_x \times \mathbf{r}_y = (0, 2y, 1). \]We check its direction, ensuring it is directed away from the \(x\)-axis.
03

Set Up the Surface Integral

The original flux integral is \(\iint_{S} \mathbf{F} \cdot \mathbf{n} \, d\sigma\). We substitute \(\mathbf{F} = z^{2} \mathbf{i} + x \mathbf{j} - 3z \mathbf{k}\) and \(\mathbf{n} = (0, 2y, 1)\). This gives:\[ \mathbf{F} \cdot \mathbf{n} = (0)(z^2) + (2y)(x) + (-3z)(1) = 2xy - 3z. \]
04

Replace Variables in the Integral

Substitute the parametric variables into \(\mathbf{F} \cdot \mathbf{n}\):\[ 2xy - 3z = 2xy - 3(4-y^2) = 2xy - 12 + 3y^2. \]Thus, the integral becomes: \[ \iint_{D} (2xy + 3y^2 - 12) \, dy \, dx, \]where \(D\) is the region \(0 \leq x \leq 1\) and \(-2 \leq y \leq 2\).
05

Evaluate the Integral Over the Region

First integrate with respect to \(y\):\[ \int_{-2}^{2} (2xy + 3y^2 - 12) \, dy = \left[ xy^2 + y^3 - 12y \right]_{-2}^{2}. \]Evaluating the boundaries,At \(y = 2\): \(4x + 8 - 24 = 4x - 16\),At \(y = -2\): \(4x - 8 + 24 = 4x + 16\).Subtract to get:\[ (4x - 16) - (4x + 16) = -32. \]
06

Complete the Flux Calculation

Now integrate with respect to \(x\):\[ \int_{0}^{1} -32 \, dx = [-32x]_{0}^{1} = -32(1) + 32(0) = -32. \]Thus, the total flux through the surface is \(-32\).

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Key Concepts

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

Parametrization
Parametrization is a method of expressing a surface or curve over a given domain using one or more parameters. This is particularly handy when calculating surface integrals or other forms of analysis on complex surfaces.
To parametrize a surface, you express each point on the surface as a function of a parameter or parameters. In the case of our exercise, we had to parametrize the parabolic cylinder defined by the equation \[ z = 4 - y^2 \]. We did this using two parameters, \(x\) and \(y\), which led to the parametrization:
  • \(\mathbf{r}(x, y) = (x, y, 4-y^2)\)
For this exercise, the parameter \((x, y)\) varies over a region defined by the inequalities \(0 \leq x \leq 1\) and \(-2 \leq y \leq 2\).
Parametrization converts a complex surface problem into a more accessible form by re-expressing the coordinates in terms of another set of variables, making otherwise difficult calculations feasible.
Normal vector
The normal vector of a surface is a vector that is perpendicular to the tangent plane of the surface at a given point. It is essential in calculating flux because it determines the orientation of the surface element.
For a parametrized surface, the partial derivatives of the parametrization can be used to find vectors tangent to the surface. In our exercise, by taking the partial derivatives \(\mathbf{r}_x\) and \(\mathbf{r}_y\), we obtained:
  • \(\mathbf{r}_x = (1, 0, 0)\)
  • \(\mathbf{r}_y = (0, 1, -2y)\)
Taking the cross product of these tangent vectors gives us the normal vector:
  • \(\mathbf{n} = \mathbf{r}_x \times \mathbf{r}_y = (0, 2y, 1)\)
The direction of this normal vector is checked to ensure it is in line with the specified direction, which in this exercise is away from the \(x\)-axis.
Surface integral
The surface integral is a concept fundamental to calculus where the integral is taken over a surface in three-dimensional space. It can be thought of as a double integral that sums over each point on that surface.
For our exercise, specifically, the surface integral we need to solve is of the form:\[ \iint_{S} \mathbf{F} \cdot \mathbf{n} \, d\sigma \]where \(\mathbf{F}\) is a vector field and \(\mathbf{n}\) is the normal vector of the surface. This integral calculates the flux, which represents how much of the vector field \(\mathbf{F}\) "flows" through the surface.
To convert this into a more feasible calculation, we express this integral in terms of the surface's parametrization, replacing \(z\) and \(d\sigma\) with their parameter equivalents, and simplify accordingly.
Parabolic cylinder
A parabolic cylinder in mathematics is a type of quadratic surface that resembles a parabola extruded along a parallel axis, creating a surface that spans infinitely in that direction.
Mathematically speaking, a parabolic cylinder can be represented as \[ z = c - y^2 \] with various constants or parameters. In our exercise, the parabolic cylinder is given by \[ z = 4 - y^2 \]. Such surfaces are noteworthy because they display parabolic cross-sections that can influence calculation strategies for surface integrals and flux.
When working with parabolic cylinders, understanding their geometry can help in managing the integrals and making sense of the parametrization, which often involves re-expressing variables as seen in the given exercise. This knowledge aids in manipulating the equations to achieve simpler forms for problem-solving.

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Most popular questions from this chapter

Gravitational field a. Find a potential function for the gravitational field $$\mathbf{F}=-G m M \frac{x \mathbf{i}+y \mathbf{j}+z \mathbf{k}}{\left(x^{2}+y^{2}+z^{2}\right)^{3 / 2}}$$ \((G, m, \text { and } M\) are constants). b. Let \(P_{1}\) and \(P_{2}\) be points at distance \(s_{1}\) and \(s_{2}\) from the origin. Show that the work done by the gravitational field in part (a) in moving a particle from \(P_{1}\) to \(P_{2}\) is $$G m M\left(\frac{1}{s_{2}}-\frac{1}{s_{1}}\right)$$

a. Show that the outward flux of the position vector field \(\mathbf{F}=\) \(x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\) through a smooth closed surface \(S\) is three times the volume of the region enclosed by the surface. b. Let \(n\) be the outward unit normal vector field on \(S\). Show that it is not possible for \(\mathbf{F}\) to be orthogonal to \(\mathbf{n}\) at every point of \(S\)

Work and area Suppose that \(f(t)\) is differentiable and positive for \(a \leq t \leq b .\) Let \(C\) be the path \(\mathbf{r}(t)=t \mathbf{i}+f(t) \mathbf{j}, a \leq t \leq b\) and \(\mathbf{F}=\) yi. Is there any relation between the value of the work integral $$ \int_{C} \mathbf{F} \cdot d \mathbf{r} $$ and the area of the region bounded by the \(t\) -axis, the graph of \(f\) and the lines \(t=a\) and \(t=b ?\) Give reasons for your answer.

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 \(x^{2}+y^{2}+\) \(z^{2}=4\)

Use a CAS to perform the following steps for finding the work done by force F over the given path: a. Find \(d r \text { for the path } r(t)=g(t) i+h(t)\\}+k(t) \mathbf{k}\) b. Evaluate the force \(\mathbf{F}\) along the path. c. Evaluate \(\int_{\mathcal{C}} \mathbf{F} \cdot d \mathbf{r}\) $$\begin{array}{l} \mathbf{F}=\left(x^{2} y\right) \mathbf{i}+\frac{1}{3} x^{3} \mathbf{j}+x y \mathbf{k} ; \quad \mathbf{r}(t)=(\cos t) \mathbf{i}+(\sin t) \mathbf{j}+ \\ \left(2 \sin ^{2} t-1\right) \mathbf{k}, \quad 0 \leq t \leq 2 \pi \end{array}$$

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