/*! 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 56 Evaluate the integral \(\int_{... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate the integral \(\int_{C}(2 x-y) d x+(x+3 y) d y\) along the path \(C\). C: line segments from \((0,0)\) to \((0,-3)\) and \((0,-3)\) to \((2,-3)\)

Short Answer

Expert verified
The evalution of the integral along the specified path results in -9.5

Step by step solution

01

Parametrize the path

The path C consists of two line segments - from (0,0) to (0,-3) and then from (0,-3) to (2,-3). Each segment can be parametrized as follows: \nSegment1: r_1(t) = (0,-t), 0 ≤ t ≤ 3 \nSegment2: r_2(t) = (t,-3), 0 ≤ t ≤ 2
02

Compute the derivatives

Calculate the derivatives of the parametrized functions: \nr_1'(t) = (0, -1) \nr_2'(t) = (1, 0)
03

Substitute into the Integral

Substitute \(x, y, dx, dy\) into the integral and evaluate along each segment. So we have: \n\(\int_{0}^{3}[(2*0 + t)0 + (0 + 3*t)(-1)]dt + \(\int_{0}^{2}[(2*t + 3)*1 + (t - 3*3)*0]dt
04

Evaluate the Integral

Evaluate both integrals and sum the result: \n= \(-\int_{0}^{3}3t dt + \int_{0}^{2}2t dt\) \n= \(-[1.5*t^2]_0^3 + [t^2]_0^2\) \n= -13.5 + 4

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Key Concepts

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

Line Integral Evaluation
The process of evaluating a line integral involves summing up the product of a vector function with the differential element along a curve. This kind of integration takes into account not just the function's value at certain points, but how those values change as you move along the path defined by the curve.

Let's consider the example provided. To evaluate the line integral \(\int_{C}(2 x-y) d x+(x+3 y) d y\), we break down the path C into simple segments that we can work with. In this case, two line segments. The integral along each piece can be thought of as a separate problem, yet they are connected by their position along the path. We must ensure that each segment is evaluated with respect to the parametrization of the curve.
Parametrization of Curves
The parametrization of a curve is a way to describe a curve by specifying the coordinates as functions of one or more parameters. It's like providing a set of instructions or a map that tells you how to trace out the curve in space as you change the parameter values. A good parametrization will make evaluating the line integral simpler and more intuitive.

In our case, we broke the path into two segments and defined r1(t) and r2(t) as the parametric equations representing each segment of the curve. Parametrizations convert complicated paths into simpler, straight-line motions that are easier to handle. Specifically, we used the parameter t to trace each segment of the path, which then guided the evaluation of the line integrals over the defined intervals.
Fundamental Theorem of Line Integrals
The Fundamental Theorem of Line Integrals states that, under certain conditions, the line integral of a gradient field along a path depends only on the values of the function at the endpoints of the path. This remarkable theorem simplifies calculations and offers deep insights into the nature of conservative vector fields.

Although the theorem isn't directly applied in the step-by-step solution to our problem, it's vital to understand the conditions under which the theorem can be applied, as it has the power to turn a seemingly hard integral into a straightforward substitution. When dealing with vector fields that are conservative, the Fundamental Theorem of Line Integrals enables us to avoid parametrization and directly evaluate the function at the end points. However, when a vector field isn't conservative, as in the given exercise, we must return to the more traditional method of line integral evaluation through parametrization and careful integration.

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Most popular questions from this chapter

Evaluate \(\int_{S} \int f(x, y, z) d S .\) \(f(x, y, z)=\sqrt{x^{2}+y^{2}+z^{2}}\) \(S: z=\sqrt{x^{2}+y^{2}}, \quad x^{2}+y^{2} \leq 4\)

Use the Divergence Theorem to evaluate \(\int_{S} \int \mathbf{F} \cdot \mathbf{N} d S\) and find the outward flux of \(\mathrm{F}\) through the surface of the solid bounded by the graphs of the equations. Use a computer algebra system to verify your results. $$ \begin{aligned} &\mathbf{F}(x, y, z)=x^{2} \mathbf{i}-2 x y \mathbf{j}+x y z^{2} \mathbf{k}\\\ &S: z=\sqrt{a^{2}-x^{2}-y^{2}}, z=0 \end{aligned} $$

Find the flux of \(F\) over the closed surface. (Let \(\mathbf{N}\) be the outward unit normal vector of the surface.) \(\mathbf{F}(x, y, z)=4 x y \mathbf{i}+z^{2} \mathbf{j}+y z \mathbf{k}\) \(S:\) unit cube bounded by \(x=0, x=1, y=0, y=1, z=0\), \(z=1\)

Find the area of the surface over the given region. Use a computer algebra system to verify your results. $$ \begin{aligned} &\text { The part of the cone } \mathbf{r}(u, v)=a u \cos v \mathbf{i}+a u \sin v \mathbf{j}+u \mathbf{k} \text { , }\\\ &\text { where } 0 \leq u \leq b \text { and } 0 \leq v \leq 2 \pi \end{aligned} $$

Let \(f\) be a nonnegative function such that \(f^{\prime}\) is continuous over the interval \([a, b]\). Let \(S\) be the surface of revolution formed by revolving the graph of \(f\), where \(a \leq x \leq b\), about the \(x\) -axis. Let \(x=u, y=f(u) \cos v\), and \(z=f(u) \sin v\), where \(a \leq u \leq b\) and \(0 \leq v \leq 2 \pi\). Then, \(S\) is represented parametrically by \(\mathbf{r}(u, v)=u \mathbf{i}+f(u) \cos v \mathbf{j}+f(u) \sin v \mathbf{k}\) Show that the following formulas are equivalent. Surface area \(=2 \pi \int_{a}^{b} f(x) \sqrt{1+\left[f^{\prime}(x)\right]^{2}} d x\) Surface area \(=\int_{D} \int\left\|\mathbf{r}_{u} \times \mathbf{r}_{v}\right\| d A\)

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