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Magazine Chapter 14: Problem 5
In Exercises \(5-10\), a planar curve \(C\) is given along with a surface \(f\) that is defined over \(C\). Evaluate the line integral \(\int_{C} f(s) d s\) \(C\) is the line segment joining the points (-2,-1) and (1,2)\(;\) the surface is \(f(x, y)=x^{2}+y^{2}+2\)
Short Answer
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Step by step solution
01
Parameterize the Line Segment
First, we need to express the line segment joining (-2,-1) and (1,2) in a parametric form. Let the parameter be \(t\) such that \(0 \leq t \leq 1\). The parametric equations are:\[ x(t) = -2 + 3t, \quad y(t) = -1 + 3t \]
02
Compute the Differential Arc Length
The differential arc length \(ds\) along the parameterized path can be expressed in terms of the parameter \(t\):\[ ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \]\( \frac{dx}{dt} = 3 \) and \( \frac{dy}{dt} = 3 \), so:\[ ds = \sqrt{3^2 + 3^2} \, dt = 3\sqrt{2} \, dt \]
03
Substitute into Line Integral
Substitute the parameterized coordinates and the arc length differential into the integral:\[ \int_{C} f(s) \, ds = \int_{0}^{1} f(x(t), y(t)) \, ds \]Since \(f(x,y) = x^2 + y^2 + 2\), we substitute \(x(t) = -2+3t\) and \(y(t) = -1+3t\) into \(f(x,y):\)\[ f(x(t), y(t)) = (-2+3t)^2 + (-1+3t)^2 + 2 \]
04
Simplify the Function Inside the Integral
Simplify the function:\[ = ((-2+3t)^2 + (-1+3t)^2 + 2) \]\[ = (4 - 12t + 9t^2) + (1 - 6t + 9t^2) + 2 \]\[ = 4 + 1 + 2 - 12t - 6t + 18t^2 \]\[ = 7 - 18t + 18t^2 \]
05
Evaluate the Integral
Now evaluate the integral:\[ \int_{0}^{1} (7 - 18t + 18t^2) \, 3\sqrt{2} \, dt \]Factor out the constant:\[ 3\sqrt{2} \int_{0}^{1} (7 - 18t + 18t^2) \, dt \]Evaluate each term separately:\[ = 3\sqrt{2} \left[7t - 9t^2 + 6t^3 \right]_{0}^{1} \]\[ = 3\sqrt{2} \left[(7 \times 1 - 9 \times 1^2 + 6 \times 1^3) - (7 \times 0 - 9 \times 0^2 + 6 \times 0^3)\right] \]\[ = 3\sqrt{2} \left[7 - 9 + 6\right] \]\[ = 3\sqrt{2} \times 4 \]\[ = 12\sqrt{2} \]
06
Final Result
The value of the line integral \(\int_{C} f(s) \, ds\) over the line segment is \(12\sqrt{2} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Parametrization of Curves
Parametrization is a way to represent a curve using a parameter, typically denoted as \( t \). This process simplifies complex geometrical shapes into more manageable forms using equations. For a line segment, such as the one connecting the points (-2,-1) and (1,2), we use \( t \) to express both \( x \) and \( y \) coordinates in terms of \( t \).
\[ x(t) = -2 + 3t, \quad y(t) = -1 + 3t, \quad 0 \leq t \leq 1 \]
Here, as \( t \) progresses from 0 to 1, \( x(t) \) and \( y(t) \) trace the segment from the initial point (-2,-1) to the terminal point (1,2). This method allows us to easily compute line integrals and understand the curve's behavior.
\[ x(t) = -2 + 3t, \quad y(t) = -1 + 3t, \quad 0 \leq t \leq 1 \]
Here, as \( t \) progresses from 0 to 1, \( x(t) \) and \( y(t) \) trace the segment from the initial point (-2,-1) to the terminal point (1,2). This method allows us to easily compute line integrals and understand the curve's behavior.
Differential Arc Length
The differential arc length, \( ds \), is a small segment of the curve; it acts like an infinitesimally small piece we can sum to find the curve's total length.
The formula for \( ds \) when considering a parameterized curve is:
\[ ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \]
For our parametrized line where \( \frac{dx}{dt} = 3 \) and \( \frac{dy}{dt} = 3 \), the differential arc length becomes:
\[ ds = \sqrt{3^2 + 3^2} \, dt = 3\sqrt{2} \, dt \]
This constant value indicates that the curve has a uniform slope, meaning the line is straight.
The formula for \( ds \) when considering a parameterized curve is:
\[ ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \]
For our parametrized line where \( \frac{dx}{dt} = 3 \) and \( \frac{dy}{dt} = 3 \), the differential arc length becomes:
\[ ds = \sqrt{3^2 + 3^2} \, dt = 3\sqrt{2} \, dt \]
This constant value indicates that the curve has a uniform slope, meaning the line is straight.
Evaluation of Integrals
Evaluating a line integral involves integrating a function over a curve. First substitute the parameterized formulas of \( x(t) \) and \( y(t) \) into the given function \( f(x, y) \). In this case, the function \( f(x, y) = x^2 + y^2 + 2 \) becomes:
\[ f(x(t), y(t)) = (-2+3t)^2 + (-1+3t)^2 + 2 \]
Simplifying this expression results in \( 7 - 18t + 18t^2 \). Substitute this back into the line integral with the expression for \( ds \):
\[ \int_{0}^{1} (7 - 18t + 18t^2) \, 3\sqrt{2} \, dt \]
Each part of the expression is integrated separately, where factors are simplified, and constants are factored out, leading to a simplified calculation.
\[ f(x(t), y(t)) = (-2+3t)^2 + (-1+3t)^2 + 2 \]
Simplifying this expression results in \( 7 - 18t + 18t^2 \). Substitute this back into the line integral with the expression for \( ds \):
\[ \int_{0}^{1} (7 - 18t + 18t^2) \, 3\sqrt{2} \, dt \]
Each part of the expression is integrated separately, where factors are simplified, and constants are factored out, leading to a simplified calculation.
Functions of Two Variables
In multivariable calculus, a function of two variables, like \( f(x, y) = x^2 + y^2 + 2 \), maps two input variables into a single output. This particular function defines a surface in the 3D coordinate system.
Understanding these functions is crucial, especially when dealing with line integrals, as they allow us to evaluate how a surface behaves over a path or specific domain.
The functions vary depending on the specific values of \( x \) and \( y \), affecting the output. In this exercise, the function describes a surface which, over the parametric path, confines the integral evaluations.
Understanding these functions is crucial, especially when dealing with line integrals, as they allow us to evaluate how a surface behaves over a path or specific domain.
The functions vary depending on the specific values of \( x \) and \( y \), affecting the output. In this exercise, the function describes a surface which, over the parametric path, confines the integral evaluations.
