Chapter 14: Problem 6
\(\int_{C}\left(x^{2}+y^{2}+z^{2}\right) d s ; \quad C\) is the curve \(x=4 \cos t\), \(y=4 \sin t, z=3 t, 0 \leq t \leq 2 \pi\).
Short Answer
Expert verified
The result of the line integral is \(160\pi + 40\pi^3\).
Step by step solution
01
Parameterize the curve
Given the parametric equations, the curve C is parameterized as \( \vec{r}(t) = \langle 4\cos t, 4\sin t, 3t \rangle \) where \( 0 \leq t \leq 2\pi \). This represents a helical curve.
02
Find the derivative of the parameterization
Calculate the derivative of \( \vec{r}(t) \) with respect to \( t \). We have \( \vec{r}'(t) = \frac{d}{dt} \langle 4\cos t, 4\sin t, 3t \rangle = \langle -4\sin t, 4\cos t, 3 \rangle \).
03
Compute the magnitude of the derivative
Find the magnitude of \( \vec{r}'(t) \): \(\left| \vec{r}'(t) \right| = \sqrt{(-4\sin t)^2 + (4\cos t)^2 + 3^2} = \sqrt{16\sin^2 t + 16\cos^2 t + 9}\). Consider that \(\sin^2 t + \cos^2 t = 1\). Thus, \( \left| \vec{r}'(t) \right| = \sqrt{25} = 5 \).
04
Set up the integral for the line integral
The integral becomes \( \int_{0}^{2\pi} (x^2 + y^2 + z^2) \left| \vec{r}'(t) \right| \ dt \). Substitute \(x=4\cos t\), \(y=4\sin t\), and \(z=3t\) in \(x^2 + y^2 + z^2\). This gives \( (4\cos t)^2 + (4\sin t)^2 + (3t)^2 = 16 + 9t^2 \). The integral becomes \( 5 \int_{0}^{2\pi} (16 + 9t^2) \ dt \).
05
Evaluate the integral
Evaluate \( 5 \int_{0}^{2\pi} (16 + 9t^2) \ dt \). This splits into two integrals: \(5 \left[ 16 \int_{0}^{2\pi} dt + 9 \int_{0}^{2\pi} t^2 \ dt \right] \). Calculate each separately: \(16t\) evaluated from 0 to \(2\pi\) is \( 32\pi \), and \( t^3/3 \) evaluated from 0 to \(2\pi\) is \(\frac{8\pi^3}{3}\). Therefore, the integral equals \(5(32\pi + 3 \times \frac{8\pi^3}{3}) = 160\pi + 40\pi^3\).
06
Simplify the result
Multiply to get the final result: \( 160\pi + 40\pi^3 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Parametric Equation
In mathematics, parametric equations offer a dynamic way to describe lines or curves through a set of equations that express the coordinates of the curve as functions of a parameter. In our case, the curve \( C \) is defined using parametric equations for \( x \), \( y \), and \( z \) based on the parameter \( t \), specifically:
- \( x = 4\cos t \)
- \( y = 4\sin t \)
- \( z = 3t \)
Vector Calculus
Vector calculus extends the ideas of standard calculus to fields that describe vector quantities. These can include vector fields where each point in space is linked not to a single value, but to a vector.
Our task involves using vector calculus to compute a line integral over a curve, \( C \), parameterized by \( \vec{r}(t) = \langle 4\cos t, 4\sin t, 3t \rangle \). A line integral like this is integral to understanding physical concepts such as work done by a force field.
To progress, compute the derivative \( \vec{r}'(t) \) which gives a vector that is tangent to the curve at each point:
Our task involves using vector calculus to compute a line integral over a curve, \( C \), parameterized by \( \vec{r}(t) = \langle 4\cos t, 4\sin t, 3t \rangle \). A line integral like this is integral to understanding physical concepts such as work done by a force field.
To progress, compute the derivative \( \vec{r}'(t) \) which gives a vector that is tangent to the curve at each point:
- \( \vec{r}'(t) = \langle -4\sin t, 4\cos t, 3 \rangle \)
Helical Curve
Helical curves are a type of three-dimensional spiral. In this instance, our parametric equations define a helix via \( x = 4\cos t \), \( y = 4\sin t \), and \( z = 3t \). These represent the characteristics of a helix, where:
- The \( x \) and \( y \) equations describe a circle of radius 4 in the plane \( z = 0 \).
- The \( z \) component, changing linearly with \( t \), adds vertical motion to the circular base.
Integral Evaluation
The evaluation of line integrals is a key skill in mathematical and engineering contexts. Starting from \( \int_{C}(x^{2} + y^{2} + z^{2})\ ds \), you compute the value by substituting the parametric functions. After substituting, simplify to find \( x^{2} + y^{2} + z^{2} = 16 + 9t^{2} \).
Then, multiply by the magnitude of the derivative, \( \left| \vec{r}'(t) \right| = 5 \), transforming it into a standard integral:
Then, multiply by the magnitude of the derivative, \( \left| \vec{r}'(t) \right| = 5 \), transforming it into a standard integral:
- \( 5 \int_{0}^{2\pi} (16 + 9t^{2}) \ dt \)
- The integral of a constant: \( \int_{0}^{2\pi} 16 \ dt = 32\pi \)
- The quadratic integral: \( \int_{0}^{2\pi} 9t^2 \ dt = \frac{8\pi^3}{3} \)