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Chapter 9: Problem 35

$$\begin{array}{l} \mathbf{r}=3 \cos t \mathbf{i}+3 \sin t \mathbf{j}, 0 \leq t \leq 2 \pi ; \quad d \mathbf{r}=-3 \sin t \mathbf{i}+3 \cos t \mathbf{j} ; \mathbf{F}=a \mathbf{i}+b \mathbf{j} \\ W=\int_{C} \mathbf{F} \cdot d \mathbf{r}=\int_{0}^{2 \pi}(-3 a \sin t+3 b \cos t) d t=\left.(3 a \cos t+3 b \sin t)\right|_{0} ^{2 \pi}=0 \end{array}$$

Short Answer

Expert verified
The work done \( W \) is zero.

Step by step solution

01

Understand the Problem

We are given a vector field \( \mathbf{F} = a \mathbf{i} + b \mathbf{j} \) and a parametrization of a curve \( \mathbf{r} = 3 \cos t \mathbf{i} + 3 \sin t \mathbf{j} \). We need to compute the line integral \( W = \int_{C} \mathbf{F} \cdot d\mathbf{r} \).
02

Compute the Derivative of \( \mathbf{r}(t) \)

The derivative of \( \mathbf{r}(t) \) gives the differential vector \( d\mathbf{r} \). Given \( \mathbf{r} = 3 \cos t \mathbf{i} + 3 \sin t \mathbf{j} \), differentiate with respect to \( t \) to get \( d\mathbf{r} = -3 \sin t \mathbf{i} + 3 \cos t \mathbf{j} \).
03

Set Up the Integral

To find the work done \( W \), set up the line integral \( W = \int_{C} \mathbf{F} \cdot d\mathbf{r} \). Substitute \( \mathbf{F} = a \mathbf{i} + b \mathbf{j} \) and \( d\mathbf{r} = -3 \sin t \mathbf{i} + 3 \cos t \mathbf{j} \).
04

Compute the Dot Product

Calculate the dot product \( \mathbf{F} \cdot d\mathbf{r} = (a \mathbf{i} + b \mathbf{j}) \cdot (-3 \sin t \mathbf{i} + 3 \cos t \mathbf{j}) = -3a \sin t + 3b \cos t \).
05

Integrate the Expression

Integrate the expression \( \int_{0}^{2\pi} (-3a \sin t + 3b \cos t) \, dt \). The integral can be separated, resulting in \( 3a \int_{0}^{2\pi} \cos t \, dt + 3b \int_{0}^{2\pi} \sin t \, dt \).
06

Solve the Indefinite Integrals

Integrate \( \int \cos t \, dt = \sin t + C \) and \( \int \sin t \, dt = -\cos t + C \).
07

Evaluate the Definite Integrals

Evaluate \( 3a [\sin t]_{0}^{2\pi} + 3b [-\cos t]_{0}^{2\pi} \) which leads to \( 3a(0) + 3b(0) = 0 \), since both sine and cosine complete full cycles.
08

Conclude the Result

Since the integral evaluated to zero, the work done by the force \( \mathbf{F} \) over the closed path is zero, meaning no net work is done.

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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 mathematical terms, if you have a vector field \( \mathbf{F} = a \mathbf{i} + b \mathbf{j} \), each part (\( a \) and \( b \)) varies with respect to the point in space it describes.
It essentially tells us both the direction and the magnitude of a quantity at every point in the space it covers.
  • **Direction**: Indicates the way the vector points at a particular spot.
  • **Magnitude**: Quantifies the size or strength of the vector at that spot.
Vector fields are incredibly useful, especially when working with physical concepts like gravitational, electrical, and magnetic fields, as they allow us to visualize forces or velocities in multidimensional space.
Parametrization of a Curve
The parametrization of a curve involves expressing the coordinates of points on the curve as functions of a parameter, often denoted as \( t \). This aids in traversing the curve by systematically going through all its points by varying \( t \).
In the given problem, the curve is parametrized by \( \mathbf{r} = 3 \cos t \mathbf{i} + 3 \sin t \mathbf{j} \), where \( t \) ranges from \( 0 \) to \( 2\pi \).

  • **Cosine Function** \( 3 \cos t \mathbf{i} \): Describes circular horizontal movement.
  • **Sine Function** \( 3 \sin t \mathbf{j} \): Describes circular vertical movement.
  • **Range** \( 0 \leq t \leq 2\pi \): Indicates a full loop of the circle, creating a closed curve.
This form of parametrization is particularly useful in scenarios involving circular paths, enabling easy calculation and analysis of motion or changes over the curve.
Dot Product
The dot product is a crucial operation in vector mathematics. It takes two vectors and returns a scalar, representing how much one vector extends in the direction of another.
The dot product \( \mathbf{F} \cdot d\mathbf{r} \) calculates the integral in the original problem:
\[(a \mathbf{i} + b \mathbf{j}) \cdot (-3 \sin t \mathbf{i} + 3 \cos t \mathbf{j}) = -3a \sin t + 3b \cos t \]
  • **Length of Projection**: The result of a dot product can be viewed as the length of the projection of one vector onto another.
  • **Geometric Interpretation**: It measures the angle between vectors; if the dot product is zero, vectors are orthogonal.
In the context of line integrals, the dot product allows us to determine the amount of force applied along the direction of the path.
Definite Integral
The definite integral is a fundamental concept in calculus that computes the net area under a curve, between two specific limits. It is represented by \( \int_{a}^{b} f(t) \, dt \), where \( a \) and \( b \) are the limits of integration.

In this exercise, the definite integral allows us to calculate the total work done over a closed path:
\[\int_{0}^{2\pi} (-3a \sin t + 3b \cos t) \, dt \]
By evaluating this integral from \( 0 \) to \( 2\pi \), we are considering the complete traversal of the circular path described by the parametrization. This evaluates to zero, indicating no net work is done for a full loop.
  • **Limits** \( \int_{a}^{b} \): Define the start and end of the evaluation interval.
  • **Net Area**: Measures the accumulated value between two points, hence explaining the total effect or change.
Definite integrals are widely applied, from calculating areas and volumes to solving real-world physics problems involving continuous quantities.

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Most popular questions from this chapter

The equations are \(\phi=\pi / 4\) and \(\rho=3\). $$\begin{aligned} V &=\int_{0}^{2 \pi} \int_{0}^{\pi / 4} \int_{0}^{3} \rho^{2} \sin \phi d \rho d \phi d \theta=\left.\int_{0}^{2 \pi} \int_{0}^{\pi / 4} \frac{1}{3} \rho^{3} \sin \phi\right|_{0} ^{3} d \phi d \theta=\int_{0}^{2 \pi} \int_{0}^{\pi / 4} 9 \sin \phi d \phi d \theta \\ &=\int_{0}^{2 \pi}-\left.9 \cos \phi\right|_{0} ^{\pi / 4} d \theta=-9 \int_{0}^{2 \pi}\left(\frac{\sqrt{2}}{2}-1\right) d \theta=9 \pi(2-\sqrt{2}) \end{aligned}$$

The equation is \(z=r\) and the density is \(\rho=k r\). $$\begin{aligned} I_{x} &=\int_{0}^{2 \pi} \int_{0}^{1} \int_{r}^{1}\left(y^{2}+z^{2}\right)(k r) r d z d r d \theta=k \int_{0}^{2 \pi} \int_{0}^{1} \int_{r}^{1}\left(r^{4} \sin ^{2} \theta+r^{2} z^{2}\right) d z d r d \theta \\ &=\left.k \int_{0}^{2 \pi} \int_{0}^{1}\left[\left(r^{4} \sin ^{2} \theta\right) z+\frac{1}{3} r^{2} z^{3}\right]\right|_{r} ^{1} d r d \theta \\\ &=k \int_{0}^{2 \pi} \int_{0}^{1}\left(r^{4} \sin ^{2} \theta+\frac{1}{3} r^{2}-r^{5} \sin ^{2} \theta-\frac{1}{3} r^{5}\right) d r d \theta \\ &=\left.k \int_{0}^{2 \pi}\left(\frac{1}{5} r^{5} \sin ^{2} \theta+\frac{1}{9} r^{3}-\frac{1}{6} r^{6} \sin ^{2} \theta-\frac{1}{18} r^{6}\right)\right|_{0} ^{1} d \theta=k \int_{0}^{2 \pi}\left(\frac{1}{30} \sin ^{2} \theta+\frac{1}{18}\right) d \theta \\ &=\left.k\left(\frac{1}{60} \theta-\frac{1}{120} \sin 2 \theta+\frac{1}{18} \theta\right)\right|_{0} ^{2 \pi}=\frac{13}{90} \pi k \end{aligned}$$

\(\int_{0}^{1} \int_{x}^{1} x^{2} \sqrt{1+y^{4}} d y d x=\int_{0}^{1} \int_{0}^{y} x^{2} \sqrt{1+y^{4}} d x d y=\left.\int_{0}^{1} \frac{1}{3} x^{3} \sqrt{1+y^{4}}\right|_{0} ^{y} d y\) \(=\frac{1}{3} \int_{0}^{1} y^{3} \sqrt{1+y^{4}} d y=\left.\frac{1}{3}\left[\frac{1}{6}\left(1+y^{4}\right)^{3 / 2}\right]\right|_{0} ^{1}=\frac{1}{18}(2 \sqrt{2}-1)\)

\(\operatorname{div} \mathbf{F}=\frac{1}{x^{2}+y^{2}+z^{2}} .\) Using spherical coordinates $$\begin{aligned}\iint_{S} \mathbf{F} \cdot \mathbf{n} d S &=\iiint_{D} \operatorname{div} \mathbf{F} d V=\int_{0}^{2 \pi} \int_{0}^{\pi} \int_{a}^{b} \frac{1}{\rho^{2}} \rho^{2} \sin \phi d \rho d \phi d \theta \\ &=\int_{0}^{2 \pi} \int_{0}^{\pi}(b-a) \sin \phi d \phi d \theta=(b-a) \int_{0}^{2 \pi}-\left.\cos \phi\right|_{0} ^{\pi} d \theta \\ &=(b-a) \int_{0}^{2 \pi} 2 d \theta=4 \pi(b-a)\end{aligned}$$

Substituting the first equation into the second, we see that the surfaces intersect in the plane \(y=4 .\) Using polar coordinates in the \(x z\) -plane, the equations of the surfaces become \(y=r^{2}\) and \(y=\frac{1}{2} r^{2}+2\). $$\begin{aligned} V &=\int_{0}^{2 \pi} \int_{0}^{2} \int_{r^{2}}^{r^{2} / 2+2} r d y d r d \theta=\int_{0}^{2 \pi} \int_{0}^{2} r\left(\frac{r^{2}}{2}+2-r^{2}\right) d r d \theta \\ &=\int_{0}^{2 \pi} \int_{0}^{2}\left(2 r-\frac{1}{2} r^{3}\right) d r d \theta=\left.\int_{0}^{2 \pi}\left(r^{2}-\frac{1}{8} r^{4}\right)\right|_{0} ^{2} d \theta=\int_{0}^{2 \pi} 2 d \theta=4 \pi \end{aligned}$$
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