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Chapter 16: Problem 12

\(\begin{array}{l}{\text { Evaluate } \int_{C} \sqrt{x^{2}+y^{2}} d s \text { along the curve } \mathbf{r}(t)=(4 \cos t) \mathbf{i}+} \\ {(4 \sin t) \mathbf{j}+3 t \mathbf{k},-2 \pi \leq t \leq 2 \pi}\end{array} \)

Short Answer

Expert verified
The value of the line integral is \(80\pi\).

Step by step solution

01

Understanding the Problem

We are asked to evaluate the line integral \( \int_{C} \sqrt{x^2 + y^2} \, ds \) where the curve \( C \) is given by \( \mathbf{r}(t) = (4 \cos t) \mathbf{i} + (4 \sin t) \mathbf{j} + 3t \mathbf{k} \) and \(-2\pi \leq t \leq 2\pi\). This curve represents a helix with a circular base in the \(xy\)-plane.
02

Parameterizing the Curve

The curve \( \mathbf{r}(t) \) is already given in parametric form with \( x = 4 \cos t \), \( y = 4 \sin t \), and \( z = 3t \). The limits of integration for \( t \) are from \(-2\pi\) to \(2\pi\).
03

Finding ds - The Arc Length Differential

To find \( ds \), we calculate the magnitude of \( \frac{d\mathbf{r}}{dt} \). First, compute \( \frac{d\mathbf{r}}{dt} = (-4\sin t) \mathbf{i} + (4\cos t) \mathbf{j} + 3 \mathbf{k} \). Then, \( ds = \left| \frac{d\mathbf{r}}{dt} \right| dt = \sqrt{(-4\sin t)^2 + (4\cos t)^2 + 3^2} \, dt = \sqrt{16 + 9} \, dt = 5 \, dt \).
04

Simplifying the Integrand

For the integrand \( \sqrt{x^2 + y^2} \), substitute \( x = 4\cos t \) and \( y = 4\sin t \). Then, \( \sqrt{x^2 + y^2} = \sqrt{(4\cos t)^2 + (4\sin t)^2} = \sqrt{16}\), which simplifies to 4.
05

Evaluating the Integral

Substitute \( \sqrt{x^2+y^2} = 4 \) and \( ds = 5 \, dt \) into the integral: \( \int_{-2\pi}^{2\pi} 4 \times 5 \, dt = 20 \int_{-2\pi}^{2\pi} \, dt \). Evaluate the integral: \( 20 \times [t]_{-2\pi}^{2\pi} = 20 \times (2\pi - (-2\pi)) = 80\pi \).

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Key Concepts

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

Helix
A helix is a fascinating three-dimensional curve that spirals around an axis, similar to the shape of a spring or a screw thread. It's characterized by a constant radius and pitch, which is the distance between subsequent turns.

In the context of the given problem, the curve \( \mathbf{r}(t) = (4 \cos t) \mathbf{i} + (4 \sin t) \mathbf{j} + 3t \mathbf{k} \) represents a helix.

Here's why:
  • The expressions \( x = 4\cos t \) and \( y = 4\sin t \) together describe a circular motion in the \( xy \)-plane with a radius of 4.
  • The term \( z = 3t \) indicates a linear progression along the \( z \)-axis.
  • As \( t \) varies, the point on the helix moves in circles in the \( xy \)-plane while concurrently progressing in the \( z \)-direction.
A helix is a very common natural form, appearing in DNA structures and even in the electromagnetism of charged particles. Understanding its geometry can deeply enhance our grasp of three-dimensional spaces.
Arc Length Differential
The arc length differential, denoted as \( ds \), is an essential concept when working with line integrals. It captures the small segment of length along a curve. Calculating \( ds \) is crucial for evaluating integrals over curves.

To find the arc length differential \( ds \) for a parametric curve \( \mathbf{r}(t) \), we need the magnitude of the derivative \( \frac{d\mathbf{r}}{dt} \):
  • In our problem, \( \frac{d\mathbf{r}}{dt} = (-4\sin t) \mathbf{i} + (4\cos t) \mathbf{j} + 3 \mathbf{k} \).
  • Calculating the magnitude: \[\left| \frac{d\mathbf{r}}{dt} \right| = \sqrt{(-4\sin t)^2 + (4\cos t)^2 + 3^2} = \sqrt{16\sin^2 t + 16\cos^2 t + 9} = \sqrt{16 + 9} = 5.\]
  • Thus, \( ds = 5\, dt \).
This magnitude captures how the curve's length changes with respect to the parameter \( t \). It allows us to translate between the parameter \( t \) and the actual length along the curve, an essential step in solving line integrals.
Parametric Equation
Parametric equations define curves by expressing coordinates as functions of a parameter, typically noted as \( t \). This approach is particularly effective for representing complex curves that cannot be described by a single function relationship between coordinates like \( y = f(x) \).

Let's unpack the parametric representation used in the problem:
  • The equation is \( \mathbf{r}(t) = (4\cos t) \mathbf{i} + (4\sin t) \mathbf{j} + 3t \mathbf{k} \).
  • Here, \( x(t) = 4\cos t \) and \( y(t) = 4\sin t \) create a circle in the 2D \( xy \)-plane with radius 4. The function \( z(t) = 3t \) adds depth, letting the curve extend into 3D, creating the helix.
  • The limits \(-2\pi \leq t \leq 2\pi\) specify how much of the curve we consider, effectively determining the helical segment's length.
These equations beautifully illustrate the movement around and up the helix as \( t \) increases. Parametric equations are powerful tools in calculus and help us explore trajectories, motion, and growth over time in real-world applications.

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Most popular questions from this chapter

In Exercises \(43-46,\) use a CAS to perform the following steps to evaluate the line integrals. $$ \begin{array}{l}{\text { a. Find } d s=|\mathbf{v}(t)| d t \text { for the path } \mathbf{r}(t)=g(t) \mathbf{i}+h(t) \mathbf{j}+k(t) \mathbf{k} \text { . }} \\ {\text { b. Express the integrand } f(g(t), h(t), k(t))|\mathbf{v}(t)| \text { as a function of }} \\ {\text { the parameter } t .} \\ {\text { c. Evaluate } \int_{C} f d s \text { using Equation }(2) \text { in the text. }}\end{array} $$ $$ \begin{array}{l}{f(x, y, z)=x \sqrt{y}-3 z^{2} ; \quad \mathbf{r}(t)=(\cos 2 t) \mathbf{i}+(\sin 2 t) \mathbf{j}+5 t \mathbf{k}} \\ {0 \leq t \leq 2 \pi}\end{array} $$

Integrate \(G(x, y, z)=x\) over the surface given by \begin{equation}z=x^{2}+y \text { for } 0 \leq x \leq 1, \quad-1 \leq y \leq 1.\end{equation}

Integrate \(G(x, y, z)=x y z\) over the triangular surface with vertices (1,0,0),(0,2,0), and (0,1,1).

Two "central" fields Find a field \(\mathbf{F}=M(x, y) \mathbf{i}+N(x, y) \mathbf{j}\) in the \(x y\) -plane with the property that at each point \((x, y) \neq(0,0), \mathbf{F}\) points toward the origin and \(|\mathbf{F}|\) is (a) the distance from \((x, y)\) to the origin, (b) inversely proportional to the distance from \((x, y)\) to the origin. (The field is undefined at \((0,0)\) .)

Regions with many holes Green's Theorem holds for a region \(R\) with any finite number of holes as long as the bounding curves are smooth, simple, and closed and we integrate over each component of the boundary in the direction that keeps \(R\) on our immediate left as we go along (see accompanying figure). $$\begin{array}{l}{\text { a. Let } f(x, y)=\ln \left(x^{2}+y^{2}\right) \text { and let } C \text { be the circle }} \\ {x^{2}+y^{2}=a^{2} . \text { Evaluate the flux integral }}\end{array}$$ $$\oint_{C} \nabla f \cdot \mathbf{n} d s$$ b. Let \(K\) be an arbitrary smooth, simple closed curve in the plane that does not pass through \((0,0) .\) Use Green's Theorem to show that $$\oint_{K} \nabla f \cdot \mathbf{n} d s$$ has two possible values, depending on whether \((0,0)\) lies inside \(K\) or outside \(K .\)
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