Problem 20 - Thomas Calculus

Question

\(\text{Evaluate}\int_{C} \sqrt{x+2 y} d s, \text { where } C \text{ is }\) $$ \begin{array}{l}{\text { a. the straight-line segment } x=t, y=4 t, \text { from }(0,0) \text { to }(1,4) \text { . }} \\ {\text { b. } C_{1} \cup C_{2} ; C_{1} \text { is the line segment from }(0,0) \text { to }(1,0) \text { and } C_{2} \text { is }} \\ {\text { the line segment from }(1,0) \text { to }(1,2) \text { . }}\end{array} $$

Step-by-Step Solution

Answer

(a) \(2\sqrt{17}\); (b) \( \frac{5\sqrt{5} + 1}{3} \)."

Step 1: Understand the Problem

We are asked to evaluate the line integral \( \int_{C} \sqrt{x+2y} \, ds \) over two different curve configurations, \( C \). The problem involves calculating the integral over specific line segments described parametrically.

Step 2: Parametrize the Curve Part (a)

For part (a), the curve \( C \) is given as the line segment where \( x = t \) and \( y = 4t \), with \( t \) ranging from \( 0 \) to \( 1 \). Essentially, this describes a straight line from the point \((0,0)\) to \((1,4)\). Substitute these into the integral and compute \( ds \).

Step 3: Calculate the Differential Element \( ds \) for Part (a)

Since \( x = t \) and \( y = 4t \), we find \( \frac{dx}{dt} = 1 \) and \( \frac{dy}{dt} = 4 \). The arc length element \( ds \) is \( \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt = \sqrt{1 + 16} \, dt = \sqrt{17} \, dt \).

Step 4: Set Up and Evaluate the Integral for Part (a)

Substitute \( x = t \) and \( y = 4t \) into \( \sqrt{x + 2y} \), which gives \( \sqrt{t + 8t} = \sqrt{9t} = 3\sqrt{t} \). The integral becomes \( \int_{0}^{1} 3\sqrt{t} \cdot \sqrt{17} \, dt = 3\sqrt{17} \int_{0}^{1} \sqrt{t} \, dt \).

Step 5: Solve the Integral for Part (a)

The integral \( \int \sqrt{t} \, dt \) is \( \frac{2}{3} t^{3/2} + C \). Evaluate from 0 to 1: \( \frac{2}{3} \cdot 1^{3/2} - \frac{2}{3} \cdot 0^{3/2} = \frac{2}{3} \). Thus, the result is \(3\sqrt{17} \cdot \frac{2}{3} = 2\sqrt{17}\).

Step 6: Parametrize the Curves for Part (b)

For \( C_1 \), the curve is from \((0,0)\) to \((1,0)\), so \( x = t \) and \( y = 0 \) where \( 0 \leq t \leq 1 \). For \( C_2 \), the curve is from \((1,0)\) to \((1,2)\), so \( x = 1 \) and \( y = t \) where \( 0 \leq t \leq 2 \).

Step 7: Evaluate the Integral over \( C_1 \)

For \( C_1: x = t, y = 0 \), the function becomes \( \sqrt{t + 2 \times 0} = \sqrt{t} \) and \( ds = dt \). The integral is \( \int_{0}^{1} \sqrt{t} \, dt = \frac{2}{3} \).

Step 8: Evaluate the Integral over \( C_2 \)

For \( C_2: x = 1, y = t \), the function becomes \( \sqrt{1 + 2t} \) and \( ds = dt \). The integral is \( \int_{0}^{2} \sqrt{1 + 2t} \, dt \). The substitution \( u = 1 + 2t \), \( du = 2 dt \), transforms the limits to \( u = 1 \) to \( u = 5 \) and \( dt = \frac{1}{2} du \). Thus, \( \int_{1}^{5} \frac{\sqrt{u}}{2} \, du \).

Step 9: Solve the Integral over \( C_2 \)

The integral \( \frac{1}{2} \int u^{1/2} \, du \) is \( \frac{1}{2} \frac{2}{3} u^{3/2} = \frac{1}{3} u^{3/2} \). Evaluate from 1 to 5: \( \frac{1}{3} (5^{3/2} - 1^{3/2}) = \frac{1}{3} (\sqrt{125} - 1) = \frac{1}{3} (5\sqrt{5} - 1) \).

Step 10: Combine Results for Part (b)

Sum the results of the integrals over \( C_1 \) and \( C_2 \): \( \frac{2}{3} + \frac{1}{3} (5\sqrt{5} - 1) = \frac{2}{3} + \frac{5\sqrt{5}}{3} - \frac{1}{3} = \frac{1}{3} + \frac{5\sqrt{5}}{3} \).

Key Concepts

Parametrization of Curves

To evaluate a line integral over a curve, the first step is to parametrize the curve, meaning we express the points on the curve as a function of a single parameter, usually denoted by \( t \). Parametrization helps us navigate from the fixed endpoints of a segment in a systematic way.
For example, in part (a) of the given exercise, the curve \( C \) is a straight line from \((0,0)\) to \((1,4)\). It is parametrized as \( x = t \) and \( y = 4t \), where \( t \) ranges from 0 to 1. This allows us to represent any point on the curve using the parameter \( t \).
In part (b), the problem involves two segments, \( C_1 \) and \( C_2 \). Parametrizing \( C_1 \), which runs from \((0,0)\) to \((1,0)\), gives us \( x = t \) and \( y = 0 \) for \( t \) in the interval \([0, 1]\). For \( C_2 \), extending from \((1,0)\) to \((1,2)\), the parametrization becomes \( x = 1 \) and \( y = t \) where \( t \) ranges from 0 to 2. By using these parametrizations, each segment of the curve is expressed explicitly, facilitating the evaluation of the line integral.

Arc Length Element

The arc length element, denoted as \( ds \), is an essential part of computing line integrals. It represents a small segment of the curve in terms of the parametrization. The formula for \( ds \) is derived by considering the Pythagorean theorem in infinitesimal terms.
For a parametric curve defined by \( x(t) \) and \( y(t) \), the arc length element is calculated as:

• \( ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \)

In part (a), since \( x = t \) and \( y = 4t \), the derivatives are \( \frac{dx}{dt} = 1 \) and \( \frac{dy}{dt} = 4 \). Thus, \( ds \) becomes \( \sqrt{1^2 + 4^2} \, dt = \sqrt{17} \, dt \).
Recognizing \( ds \) gives a tangible sense of the incremental length along the path. Without it, integrations involving line integrals wouldn't accurately account for the actual path traversed by the curve.

Integral Evaluation

After parametrizing the curve and computing the arc length element, the next step is to evaluate the line integral. In this step, you substitute the parametrizations and arc length element into the integral formula and solve.
The given line integral is \( \int_{C} \sqrt{x + 2y} \, ds \). By substituting the expressions for \( x \), \( y \), and \( ds \), the integral over \( C \) transforms into a function of \( t \), simplifying the evaluation process.
For part (a), with \( x = t \), \( y = 4t \), and \( ds = \sqrt{17} \, dt \), the expression becomes \( \sqrt{9t} = 3\sqrt{t} \). The integral \( \int_{0}^{1} 3\sqrt{t} \sqrt{17} \, dt \) evaluates to \( 2\sqrt{17} \).
Similarly, evaluate the integrals over \( C_1 \) and \( C_2 \) using their respective parametrizations and add these results to form the total for part (b). The simplification of integrals and substitutions make solving the integrals feasible—a crucial skill in applying calculus to practical problems. Integrating functions of \( t \) systematically yields the solutions for line integrals as shown in the problem.