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Chapter 6: Problem 83

For the following exercises, use a CAS to evaluate the given line integrals. [T] Evaluate \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\) where \(\mathbf{F}(x, y, z)=x^{2} y \mathbf{i}+(x-z) \mathbf{j}+x y z \mathbf{k}\) and \(C : \mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j}+2 \mathbf{k}, 0 \leq t \leq 1\)

Short Answer

Expert verified
The integral evaluates to \( -\frac{17}{15} \).

Step by step solution

01

Parametrize the Components of the Vector Field

The given vector field is \( \mathbf{F} = x^2 y \mathbf{i} + (x-z) \mathbf{j} + xy z \mathbf{k} \). Substitute the parameterization of the curve \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} + 2 \mathbf{k} \) into \( \mathbf{F} \). This gives the components as: \( x = t \), \( y = t^2 \), and \( z = 2 \). So, \( \mathbf{F}(t) = t^2 t^2 \mathbf{i} + (t-2) \mathbf{j} + t t^2 2 \mathbf{k} \) becomes \( t^4 \mathbf{i} + (t-2) \mathbf{j} + 2t^3 \mathbf{k} \).
02

Find the Derivative of the Parametrization

Find \( \frac{d\mathbf{r}}{dt} \) based on \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} + 2 \mathbf{k} \). The derivative is \( \frac{d\mathbf{r}}{dt} = \frac{d}{dt}(t \mathbf{i} + t^2 \mathbf{j} + 2 \mathbf{k}) \) which equals \( \mathbf{i} + 2t \mathbf{j} \).
03

Calculate the Dot Product

Form the dot product \( \mathbf{F}(t) \cdot \frac{d\mathbf{r}}{dt} \). Substituting in the expressions from the previous steps, we have \( (t^4 \mathbf{i} + (t-2) \mathbf{j} + 2t^3 \mathbf{k}) \cdot (\mathbf{i} + 2t \mathbf{j}) = t^4 \cdot 1 + (t-2) \cdot 2t + 2t^3 \cdot 0 \). This simplifies to \( t^4 + 2t(t-2) = t^4 + 2t^2 - 4t \).
04

Evaluate the Integral

Integrate the dot product over the given interval from 0 to 1: \( \int_0^1 (t^4 + 2t^2 - 4t) \, dt \). Calculate this integral:\[\int_0^1 t^4 \, dt + \int_0^1 2t^2 \, dt - \int_0^1 4t \, dt\]\[= \left[ \frac{t^5}{5} \right]_0^1 + 2 \left[ \frac{t^3}{3} \right]_0^1 - 4 \left[ \frac{t^2}{2} \right]_0^1\]\[ = \left( \frac{1}{5} + \frac{2}{3} - 2 \right)\]Convert all terms to have a common denominator to simplify: \( \frac{1}{5} + \frac{2}{3} \) simplified to a common denominator is \( \frac{3}{15} + \frac{10}{15} = \frac{13}{15} \). Therefore, \( \frac{13}{15} - 2 = \frac{13}{15} - \frac{30}{15} = -\frac{17}{15} \).

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Key Concepts

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

Vector Field
In mathematics and physics, a vector field assigns a vector to each point in space. It's a method to represent a region in which each point has a vector associated with it. In this exercise, the vector field is given by \( \mathbf{F}(x, y, z) = x^2 y \mathbf{i} + (x-z) \mathbf{j} + xy z \mathbf{k} \). This means:
  • The \(\mathbf{i}\) component is \(x^2y\).
  • The \(\mathbf{j}\) component is \((x-z)\).
  • The \(\mathbf{k}\) component is \(xyz\).
By substituting specific values for \(x\), \(y\), and \(z\), we can determine the vector at any point \((x, y, z)\) in the field. Understanding vector fields is crucial for solving problems involving forces, velocities, and other vector-related quantities.
Parametrization
Parametrization is a way to express a curve or surface through variables, usually called parameters. In line integrals, parametrization helps trace a path along which integration will be performed. The exercise uses the curve \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} + 2 \mathbf{k} \), where:
  • \(x = t\)
  • \(y = t^2\)
  • \(z = 2\)
The parameter \(t\) varies within the interval \([0, 1]\), determining the start and end of the curve. By substituting these components into the vector field, we obtain a new, simplified version of the field along the curve. This process is essential as it transforms the 3D space vector field into a form that can be worked with easily in integration.
Dot Product
The dot product, also known as the scalar product, is a way of multiplying two vectors that results in a scalar. It's crucial in line integrals as it finds the component of one vector along another. For the given exercise, we calculate the dot product of the vector field \(\mathbf{F}(t)\) and the derivative of the parametrization \(\frac{d\mathbf{r}}{dt}\). This is expressed as:
  • \( \mathbf{F}(t) \cdot \frac{d\mathbf{r}}{dt} = (t^4 \mathbf{i} + (t-2) \mathbf{j} + 2t^3 \mathbf{k}) \cdot (\mathbf{i} + 2t \mathbf{j}) \)
  • This results in the expression \(t^4 + 2t(t-2)\).
The dot product simplifies the integral into a single function \(t^4 + 2t^2 - 4t\), which can be readily evaluated, showing how intertwined vector operations and calculus can be in problem solving.
Integral Calculation
Calculating the integral of the dot product over a specified interval yields the value of the line integral. In this problem, we evaluate \( \int_0^1 (t^4 + 2t^2 - 4t) \, dt \). The integral is broken down into parts:
  • \( \int_0^1 t^4 \, dt = \left[ \frac{t^5}{5} \right]_0^1 \)
  • \( \int_0^1 2t^2 \, dt = 2 \left[ \frac{t^3}{3} \right]_0^1 \)
  • \( -\int_0^1 4t \, dt = -4 \left[ \frac{t^2}{2} \right]_0^1 \)
After integrating each part, we find:
  • \( \frac{1}{5} + \frac{2}{3} - 2 \)
Converting to a common denominator, we get \(-\frac{17}{15}\). Thus, the result of integrating the vector field over the parametrized curve is \(-\frac{17}{15}\). Understanding how to perform these integration steps is fundamental for evaluating complex integrals in physics and engineering.

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Most popular questions from this chapter

True or False? Line integral \(\int_{C} f(x, y) d s\) is equal to a definite integral if \(C\) is a smooth curve defined on \([a, b]\) and if function \(f\) is continuous on some region that contains curve \(C .\)

Take paraboloid \(z=x^{2}+y^{2},\) for \(0 \leq z \leq 4,\) and slice it with plane \(y=0 .\) Let \(S\) be the surface that remains for \(y \geq 0,\) including the planar surface in the \(x z\) -plane. Let \(C\) be the semicircle and line segment that bounded the cap of \(S\) in plane \(z=4\) with counterclockwise orientation. Let \(\mathbf{F}=\langle 2 z+y, 2 x+z, 2 y+x\rangle\) Evaluate \(\mathscr{J}_{S}(\nabla \times \mathbf{F}) \cdot \mathbf{n} d S\)

For the following exercises, find the flux. Let \(\mathbf{F}=5 \mathbf{i}\) and let \(C\) be curve \(y=0,0 \leq x \leq 4\) Find the flux across \(C .\)

For the following exercises, use a computer algebra system (CAS) to evaluate the line integrals over the indicated path. $$ [\mathrm{T}] \int_{C}(x-y) d s $$ $$ C : \mathbf{r}(t)=4 t \mathbf{i}+3 t \mathbf{j} \text { when } 0 \leq t \leq 2 $$

For the following exercises, find the work done. Find the mass of a wire in the shape of a circle of radius 2 centered at \((3,4)\) with linear mass density \(\rho(x, y)=y^{2}\) .
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