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Chapter 16: Problem 40

Definite integral as a line integral Suppose that a nonnegative function \(y=f(x)\) has a continuous first derivative on \([a, b] .\) Let \(C\) be the boundary of the region in the \(x y\) -plane that is bounded below by the \(x\) -axis, above by the graph of \(f,\) and on the sides by the lines \(x=a\) and \(x=b .\) Show that $$\int_{a}^{b} f(x) d x=-\oint_{C} y d x.$$

Short Answer

Expert verified
The statement is verified: \(\int_a^b f(x) \, dx = -\oint_C y \, dx\).

Step by step solution

01

Parameterize the Curve C

The curve \(C\) consists of four parts: along the curve \(y = f(x)\) from \(x = a\) to \(x = b\), downward along the line \(x = b\) from \(y = f(b)\) back to the x-axis, along the x-axis from \(x = b\) to \(x = a\), and upward along the line \(x = a\) back to \(y = f(a)\). We must parameterize each segment appropriately:
02

Parameterize the segment along y = f(x)

Consider the path \((x, f(x))\) from \(x = a\) to \(x = b\). We have \(dx = dx\) and \(dy = f'(x)dx\) for this segment.
03

Parameterize the segment x = b

As the path travels downward from \((b, f(b))\) to \((b, 0)\), parameterize this segment as \((b, y)\) for \(y\) decreasing from \(f(b)\) to \(0\). Here, \(dx = 0\) and \(dy = -dy\).
04

Parameterize the segment along x-axis

Then, move along the x-axis from \((b, 0)\) to \((a, 0)\). The path is given by \((x, 0)\) for \(x\) from \(b\) back to \(a\). Thus, \(dy = 0\) and \(dx = -dx\).
05

Parameterize the segment x = a

Finally, go upward from \((a, 0)\) to \((a, f(a))\) with path \((a, y)\) where \(y\) increases from \(0\) to \(f(a)\). The differential is \(dx = 0\) and \(dy = dy\).
06

Evaluate the Line Integral

We evaluate \(-\oint_C y \, dx\) by summing contributions from each parameterized segment. Specifically:- Segment (1): Contribution is \(-\int_a^b f(x) \, dx\).- Segment (2): Integral is 0 because \(dx = 0\).- Segment (3): Integral of \(ydx\) over an x-axis segment yields 0 since \(y = 0\).- Segment (4): Again, integral is 0 as \(dx = 0\).
07

Compile the Results

Combining results, we find that \(-\oint_C y \, dx = -\int_a^b f(x) \, dx\). Negating both sides gives the desired result: \(\int_a^b f(x) \, dx = -\oint_C y \, dx\).

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Key Concepts

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

Definite Integral
A definite integral is a fundamental concept in calculus that computes the accumulation of quantities, or the area under a curve, between two specified points. In our example, the definite integral is written as \( \int_{a}^{b} f(x) \, dx \), where \(a\) and \(b\) are the limits of integration that define the interval on the x-axis.
  • This integral measures the net area between the graph of \(f(x)\) and the x-axis over the interval \([a, b]\).
  • A positive integral value indicates that the majority of the curve is above the x-axis, while a negative value suggests it is below.
  • The calculation involves taking the function \(f(x)\) and integrating it over the interval using fundamental integration rules and techniques.
The solution shows us that the definite integral can be expressed as a line integral over a closed curve, which blends the concepts of calculus and geometry beautifully.
Parameterization
Parameterization is a powerful technique in calculus used to express curves by using a parameter, typically denoted as \( t \). This method simplifies the process of handling complex curves by transforming them into simpler, more manageable equations.
  • In our exercise, the curve \(C\) was divided into four parts and each segment was parameterized.
  • The parameterization involves setting equations for \(x\) and \(y\) in terms of \(t\), such as \( (x(t), y(t)) \).
  • This allows us to calculate integrals by expressing all necessary equations in terms of \(t\).
By using parameterization, the line integral around the curve \(C\) becomes a series of simpler integrals that are easier to evaluate, using the parameters to track movement along each piece of the curve.
Curve Integration
Curve integration, specifically line integrals, extends the concept of integrals to curves in the plane or space. Instead of integrating over a straight interval, we integrate over a path, which may be curved or segmented.
  • In the given problem, the line integral is written as \( -\oint_C y \, dx \), representing integration around the closed curve \(C\).
  • This type of integral is useful for finding quantities like work done by a force field along a path, or the circulation of a field around a loop.
  • Each segment of the curve \(C\) contributes to the total value of the line integral, highlighting the importance of careful parameterization and evaluation of each segment.
Understanding curve integration is crucial for solving problems that involve paths in vector fields and can transform a geometric problem into an analytical one, displaying the integration over a region or boundary.
Differential Equations
Differential equations are equations involving derivatives, which represent rates of change. They are essential tools for modeling continuous processes in mathematics, physics, and engineering. While the exercise focuses more on the line integral, understanding differential equations is key to interpreting dynamics involved.
  • In our scenario, calculating differentials like \(dy = f'(x) dx\) comes into play during the integration process.
  • Solving these equations helps us find functions that describe processes and predict future behavior.
  • Applications include everything from physics to finance, where they model real-world systems.
By incorporating differential equations into the integration process, one can better understand the relationship between different variables and how they change regarding the parameterized path.

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Most popular questions from this chapter

Find the area of the surfaces in Exercises \(49-54\) $$ \begin{array}{l}{\text { The triangle cut from the plane } 2 x+6 y+3 z=6 \text { by the bounding }} \\ {\text { planes of the first octant. Calculate the area three ways, using }} \\ {\text { different explicit forms. }}\end{array} $$

Green's Theorem and Laplace's equation Assuming that all the necessary derivatives exist and are continuous, show that if \(f(x, y)\) satisfies the Laplace equation $$\frac{\partial^{2} f}{\partial x^{2}}+\frac{\partial^{2} f}{\partial y^{2}}=0,$$ then $$\oint_{C} \frac{\partial f}{\partial y} d x-\frac{\partial f}{\partial x} d y=0$$ for all closed curves \(C\) to which Green's Theorem applies. (The converse is also true: If the line integral is always zero, then \(f\) satisfies the Laplace equation.)

A revealing experiment By experiment, you find that a force field \(\mathbf{F}\) performs only half as much work in moving an object along path \(C_{1}\) from \(A\) to \(B\) as it does in moving the object along path \(C_{2}\) from \(A\) to \(B\) . What can you conclude about \(\mathbf{F}\) ? Give reasons for your answer.

a. A torus of revolution (doughnut) is obtained by rotating a circle \(C\) in the \(x z\) -plane about the \(z\) -axis in space. (See the accompanying figure.) If \(C\) has radius \(r>0\) and center \((R, 0,0),\) show that a parametrization of the torus is $$ \begin{aligned} \mathbf{r}(u, \boldsymbol{v})=&((R+r \cos u) \cos v) \mathbf{i} \\ &+((R+r \cos u) \sin v) \mathbf{j}+(r \sin u) \mathbf{k} \end{aligned} $$ where \(0 \leq u \leq 2 \pi\) and \(0 \leq v \leq 2 \pi\) are the angles in the figure. b. Show that the surface area of the torus is \(A=4 \pi^{2} R r\)

Centroid Find the centroid of the portion of the sphere \(x^{2}+y^{2}+z^{2}=a^{2}\) that lies in the first octant.
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