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Chapter 16: Problem 6

I-16 Evaluate the line integral, where \(C\) is the given curve. $$\int_{C} x e^{y} d x,$$ \(C\) is the arc of the curve \(x=e^{y}\) from \((1,0)\) to \((e, 1)\)

Short Answer

Expert verified
The value of the line integral is \(\frac{1}{2}(e^2 - 1)\).

Step by step solution

01

Parameterize the Curve

Since the curve is given as \(x = e^y\), we will use \(y\) as the parameter. Let \(x(t) = e^t\) and \(y(t) = t\). The parameter \(t\) ranges from 0 to 1 in order to cover the entire curve from \((1, 0)\) to \((e, 1)\).
02

Substitute Parameters into Integral

Substitute the parameterization into the integral. The integral becomes: \[\int_{0}^{1} e^t e^t \frac{dx}{dt} dt.\] Since \(x = e^t\), \(\frac{dx}{dt} = e^t\). Thus, the integral simplifies to: \[\int_{0}^{1} (e^t)^2 dt.\]
03

Simplify the Integrand

Combine the exponential terms: \((e^t)^2 = e^{2t}\). The integral now is: \[\int_{0}^{1} e^{2t} dt.\]
04

Integrate with Respect to t

The integral of \(e^{2t}\) with respect to \(t\) is \(\frac{1}{2}e^{2t}\). Apply the limits of integration from 0 to 1: \[\left[\frac{1}{2}e^{2t}\right]_{0}^{1}.\]
05

Evaluate the Integral at Limits

Substitute the limits of integration into the evaluated expression: \[\frac{1}{2}e^{2 \cdot 1} - \frac{1}{2}e^{2 \cdot 0} = \frac{1}{2}e^2 - \frac{1}{2}e^0.\] Calculate \(e^0 = 1\), thus: \[\frac{1}{2}e^2 - \frac{1}{2}.\]
06

Simplify the Result

Factor out \(\frac{1}{2}\) to simplify the result: \[\frac{1}{2}(e^2 - 1).\] This is the final value of the line integral.

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Key Concepts

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

Parameterization
Parameterization is a technique used in integral calculus to simplify complex curves by expressing them in terms of a parameter. In this specific exercise, the curve is given by the equation \(x = e^y\). By setting \(y(t) = t\), we substitute \(y\) with parameter \(t\), which makes it easier to compute the integral along the curve.
  • We express \(x\) as \(x(t) = e^t\) and let \(y(t) = t\).
  • The parameter \(t\) ranges from 0 to 1, ensuring we follow the path from the point \((1,0)\) to \((e,1)\).

Parameterization helps transform the line integral into a single-variable calculus problem, facilitating easier computation. The process is crucial for solving line integrals since it converts a two-dimensional problem into a one-dimensional one.
Integral Calculus
Integral calculus focuses on the concept of accumulation. In the context of line integrals, it involves integrating a function over a curve rather than just a straight line on the Cartesian plane. This exercise requires evaluating a line integral, which can be daunting, especially when dealing with exponential functions.
In solving line integrals:
  • The function is parameterized to simplify the form.
  • We substitute the parameterized equations into the integral, converting the problem into one involving the independent variable \(t\).
  • By simplifying, we transform the exponentials to easier terms such as \((e^t)^2 = e^{2t}\).
  • The final task involves integrating with respect to \(t\).

In this specific exercise, the actual integration step involved the antiderivative of \(e^{2t}\), which is \(\frac{1}{2}e^{2t}\). Then applying the fundamental theorem of calculus, which involves evaluating at the bounds, gives us the result.
Exponential Function
The exponential function is a mathematical function denoted by \(e^x\), where \(e\) is a constant approximately equal to 2.71828. Exponential functions are vital in line integrals as they often describe growth and decay processes.
For this exercise:
  • The exponential function appears in both the curve \(x = e^y\) and inside the integral \(x e^y\).
  • The computation involves simplifying squared exponentials \((e^t)^2\) which becomes \(e^{2t}\).

Working with exponential functions in integrals often requires understanding their unique derivative and integration properties. The integral of \(e^{kt}\) with respect to \(t\) is \(\frac{1}{k}e^{kt}\), which was a key step in solving the integral in this exercise. Understanding these properties makes handling exponential terms during integration more manageable.

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Most popular questions from this chapter

A thin wire has the shape of the first-quadrant part of the circle with center the origin and radius a. If the density function is \(\rho(x, y)=k x y,\) find the mass and center of mass of the wire.

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Let \(\mathbf{r}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\) and \(r=|\mathbf{r}|\) \(\begin{array}{l}{\text { Verify each identity. }} \\ {\text { (a) } \nabla \cdot \mathbf{r}=3} \\ {\text { (c) } \nabla^{2} r^{3}=12 r}\end{array} \quad(b) \nabla \cdot(r \mathbf{r})=4 r\)

\(37-47\) Find the area of the surface. The part of the hyperbolic paraboloid \(z=y^{2}-x^{2}\) that lies between the cylinders \(x^{2}+y^{2}=1\) and \(x^{2}+y^{2}=4\)

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