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Special line integrals Prove the following identities, where \(C\) is a simple closed smooth oriented curve. \(\int_{C} f(x) d x+g(y) d y=0,\) where \(f\) and \(g\) have continuous derivatives on the region enclosed by \(C\)

Short Answer

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Question: Prove that the integral \(\int_{C} f(x) d x + g(y) d y = 0\) for a simple closed smooth oriented curve \(C\) and two functions \(f(x)\) and \(g(y)\) with continuous derivatives on the region enclosed by \(C\). Answer: Using Green's theorem and showing that the partial derivatives of \(f(x)\) and \(g(y)\) with respect to \(x\) and \(y\) are both zero, we proved that the integral \(\int_{C} f(x) d x + g(y) d y = 0\).

Step by step solution

01

Recall Green's theorem

Green's theorem states that if \(\textbf{F} = P(x, y) \textbf{i} + Q(x, y) \textbf{j}\) is a vector field with continuous first partial derivatives on an open region containing \(D\) where \(C\) is a simple, closed, and positively-oriented curve bounding region \(D\), then: \(\oint_{C} (P d x + Q d y) = \iint_{D} (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) dA.\) In our case, we have the vector field \(\textbf{F}(x, y) = f(x) \textbf{i} + g(y) \textbf{j}\). In order to satisfy the conditions of Green's theorem, we will first compute the partial derivatives of \(f(x)\) and \(g(y)\) to obtain the necessary form.
02

Calculate the partial derivatives of \(f(x)\) and \(g(y)\)

Since \(\textbf{F}(x, y) = f(x) \textbf{i} + g(y) \textbf{j}\), we have \(P(x, y) = f(x)\) and \(Q(x, y) = g(y)\). We need to compute the partial derivatives \(\frac{\partial P}{\partial y}\) and \(\frac{\partial Q}{\partial x}\). \(\frac{\partial P}{\partial y} = \frac{\partial f(x)}{\partial y} = 0\) because \(f(x)\) does not depend on \(y\). Similarly, \(\frac{\partial Q}{\partial x} = \frac{\partial g(y)}{\partial x} = 0\) because \(g(y)\) does not depend on \(x\).
03

Apply Green's theorem

Now that we know \(\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} = 0\), we can apply Green's theorem as follows: \(\oint_{C} (f(x) d x + g(y) d y) = \iint_{D} (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) dA = \iint_{D} (0 - 0) dA = 0\). Therefore, according to Green's theorem, the integral \(\oint_{C} f(x) d x + g(y) d y = 0\). And we have proven the given identity.

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Key Concepts

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

Line Integrals
Line integrals are a fascinating way to extend the concept of integration to curves. Instead of finding the area under a curve, we evaluate a function along a curve. This curve can be in two or three dimensions. Think of it as walking along a path and adding up values of a function at different points on the path. Line integrals are used to evaluate the work done by a force field, among other applications. In mathematical terms, a line integral of a vector field \( \textbf{F}(x, y) = P(x, y) \textbf{i} + Q(x, y) \textbf{j} \) along a curve \( C \) is given by:\[\int_{C} P \, dx + Q \, dy\]In our problem, the curve \( C \) is closed, meaning it returns to its starting point, and smooth, meaning there are no sharp turns. When applying line integrals, it's crucial to consider the orientation this path takes, as direction can impact the result.
Vector Fields
Vector fields are like imaginary maps where each point is associated with a vector (a direction and a magnitude). These fields are very common in physics where they represent quantities like velocity fields in fluids or electromagnetic fields.A vector field is usually described by a function, \( \textbf{F}(x, y) = P(x, y) \textbf{i} + Q(x, y) \textbf{j} \), where \( P \) and \( Q \) are functions of \( x \) and \( y \). In our given problem, the vector field is simple: \( \textbf{F}(x, y) = f(x) \textbf{i} + g(y) \textbf{j} \). Note how this field only depends on one variable at a time for each component, which impacts how we go about solving line integrals with it. Understanding how the vector field flows and behaves is vital when applying the line integrals since the field's nature will determine if Green’s Theorem can be applied, as it was in the solution.
Partial Derivatives
Partial derivatives are like regular derivatives but for functions with multiple variables. They show how a function changes as one of the variables changes while the other remains constant. In the context of vector fields and Green’s Theorem, partial derivatives play a critical role. For example, for the function \( P(x, y) = f(x) \), the partial derivative with respect to \( y \) is \( \frac{\partial f(x)}{\partial y} = 0 \) because \( f(x) \) does not depend on \( y \). Similarly, for \( Q(x, y) = g(y) \), the partial derivative with respect to \( x \) is \( \frac{\partial g(y)}{\partial x} = 0 \).These calculations are essential in applying Green's theorem accurately. The idea is to find the difference between these partial derivatives, which in our problem becomes zero, allowing us to conclude that the integral of the vector field over the closed curve \( C \) is zero.

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Most popular questions from this chapter

A square plate \(R=\\{(x, y):\) \(0 \leq x \leq 1,0 \leq y \leq 1\\}\) has a temperature distribution \(T(x, y)=100-50 x-25 y\) a. Sketch two level curves of the temperature in the plate. b. Find the gradient of the temperature \(\nabla T(x, y)\) c. Assume the flow of heat is given by the vector field \(\mathbf{F}=-\nabla T(x, y) .\) Compute \(\mathbf{F}\) d. Find the outward heat flux across the boundary \(\\{(x, y): x=1,0 \leq y \leq 1\\}\) e. Find the outward heat flux across the boundary \(\\{(x, y): 0 \leq x \leq 1, y=1\\}\)

Find the general formula for the surface area of a cone with height \(h\) and base radius \(a\) (excluding the base).

Surfaces of revolution Suppose \(y=f(x)\) is a continuous and positive function on \([a, b] .\) Let \(S\) be the surface generated when the graph of \(f\) on \([a, b]\) is revolved about the \(x\) -axis. a. Show that \(S\) is described parametrically by \(\mathbf{r}(u, v)=\langle u, f(u) \cos v, f(u) \sin v\rangle,\) for \(a \leq u \leq b\) \(0 \leq v \leq 2 \pi\) b. Find an integral that gives the surface area of \(S\) c. Apply the result of part (b) to the surface generated with \(f(x)=x^{3},\) for \(1 \leq x \leq 2\) d. Apply the result of part (b) to the surface generated with \(f(x)=\left(25-x^{2}\right)^{1 / 2},\) for \(3 \leq x \leq 4\)

Stokes' Theorem on closed surfaces Prove that if \(\mathbf{F}\) satisfies the conditions of Stokes' Theorem, then \(\iint_{S}(\nabla \times \mathbf{F}) \cdot \mathbf{n} d S=0\) where \(S\) is a smooth surface that encloses a region.

Ampère's Law The French physicist André-Marie Ampère \((1775-1836)\) discovered that an electrical current \(I\) in a wire produces a magnetic field \(\mathbf{B}\). A special case of Ampère's Law relates the current to the magnetic field through the equation \(\oint_{C} \mathbf{B} \cdot d \mathbf{r}=\mu I,\) where \(C\) is any closed curve through which the wire passes and \(\mu\) is a physical constant. Assume the current \(I\) is given in terms of the current density \(\mathbf{J}\) as \(I=\iint_{S} \mathbf{J} \cdot \mathbf{n} d S,\) where \(S\) is an oriented surface with \(C\) as a boundary. Use Stokes' Theorem to show that an equivalent form of Ampere's Law is \(\nabla \times \mathbf{B}=\mu \mathbf{J}\).

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