Question

Green's Theorem as a Fundamental Theorem of Calculus Show that if the flux form of Green's Theorem is applied to the vector field \(\left\langle\frac{f(x)}{c}, 0\right\rangle,\) where \(c>0\) and \(R=\\{(x, y): a \leq x \leq b, 0 \leq y \leq c\\},\) then the result is the Fundamental Theorem of Calculus, $$ \int_{a}^{b} \frac{d f}{d x} d x=f(b)-f(a) $$

Short answer

Question: Show that the flux form of Green's theorem applied to a specific vector field results in the Fundamental theorem of Calculus. Answer: By applying the flux form of Green's theorem to the given vector field F(x,y) = ⟨f(x)/c, 0⟩ and the region bounded by a ≤ x ≤ b and 0 ≤ y ≤ c, we find that the total line integral is equal to the integral of the curl of the vector field over the region. Due to the specific vector field chosen, the curl is equal to 0, leading to the line integral being equal to 0. This can be simplified to show it is equivalent to the Fundamental Theorem of Calculus.

Step-by-step solution

State the Flux Form of Green's Theorem

The flux form of Green's Theorem states that, for a continuously differentiable vector field \(\textbf{F}(x,y) = \langle P(x,y),Q(x,y) \rangle\) and a simple, positively oriented and piecewise smooth curve \(C\) enclosing a region \(D\), $$ \oint_C \textbf{F} \cdot \textbf{n} ds = \iint_D (\nabla \times \textbf{F}) dA $$ In this case, we are given the vector field \(\textbf{F}(x,y) = \left\langle\frac{f(x)}{c}, 0\right\rangle\) and the region \(R\) defined by \(a \le x \le b\) and \(0 \le y \le c\). Therefore, we need to apply the flux form of Green's Theorem to this specific case.

Calculate the Curl of the Vector Field

The curl of a vector field \(\textbf{G}(x, y) = \langle P(x,y), Q(x,y)\rangle\) is given by: $$ \nabla \times \textbf{G} = \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} $$ For our vector field, we have \(P(x,y)=\dfrac{f(x)}{c}\) and \(Q(x,y)=0\). We can then find the partial derivatives: $$ \frac{\partial P}{\partial y} = 0 \quad \text{and} \quad \frac{\partial Q}{\partial x} = 0 $$ So the curl of our vector field \(\textbf{F}(x,y)\) is: $$ \nabla \times \textbf{F} = 0 $$

Identify the Boundary Curve C of the Region R

The region R is a rectangle with vertices at \((a, 0), (a, c), (b, 0), (b, c)\). The boundary curve C consists of four line segments: 1. The bottom edge: \(C_1: \ x=a,\ 0 \le y \le c\) 2. The top edge: \(C_3: \ x=b,\ 0 \le y \le c\) 3. The left edge: \(C_2: \ a \le x \le b,\ y=0\) 4. The right edge: \(C_4: \ a \le x \le b,\ y=c\)

Calculate the Line Integral for Each Component of the Boundary Curve

Parameterize each segment of the boundary curve: 1. For \(C_1\), let \(x=a\) and \(y=y\). Then, \(ds = \dfrac{\partial y}{\partial t} dt = dy\) and \(\textbf{n} = \langle 0, 1\rangle\). The line integral is then given by: $$ \oint_{C_1} \frac{f(a)}{c} dy = 0 $$ since \(P(a,y)=0\). 2. For \(C_3\), let \(x=b\) and \(y=y\). Then, \(ds =-\dfrac{\partial y}{\partial t} dt = -dy\) and \(\textbf{n} = \langle 0, -1\rangle\). The line integral is then given by: $$ \oint_{C_3} \frac{f(b)}{c} dy = \int_{0}^{c} \frac{-f(b)}{c} dy = -\frac{f(b)}{c} \int_{0}^{c} dy $$ 3. For \(C_2\), let \(x=x\) and \(y=0\). Then, \(ds = \dfrac{\partial x}{\partial t} dt= dx\) and \(\textbf{n} = \langle -1, 0\rangle\). The line integral is then given by: $$ \oint_{C_2} 0 dx = 0 $$ since \(Q(x,0)=0\). 4. For \(C_4\), let \(x=x\) and \(y=c\). Then, \(ds =-\dfrac{\partial x}{\partial t} dt = -dx\) and \(\textbf{n} = \langle 1, 0\rangle\). The line integral is then given by: $$ \oint_{C_4} 0 dx = 0 $$ since \(Q(x,c)=0\). Therefore, the total line integral is given by \(\oint_C \textbf{F} \cdot \textbf{n} ds = -\frac{f(b)}{c} \int_{0}^{c} dy - 0 - 0 - 0\).

Apply Green's Theorem and Simplify to Obtain the Fundamental Theorem of Calculus

We can apply Green's Theorem to our problem: $$ \oint_C \textbf{F} \cdot \textbf{n} ds = \iint_R (\nabla \times \textbf{F}) dA $$ From Step 2, we know that \(\nabla \times \textbf{F} = 0\). Thus the right side of the equation becomes: $$ \iint_R 0 dA = 0 $$ Now, the left side of the equation is given by the line integral calculated in Step 4: $$ \oint_C \textbf{F} \cdot \textbf{n} ds = -\frac{f(b)}{c} \int_{0}^{c} dy $$ Equating both sides, we have: $$ -\frac{f(b)}{c} \int_{0}^{c} dy = 0 $$ Divide both sides by \(c\) and multiply both sides by -1 to get: $$ \int_{a}^{b} \frac{d f}{d x} d x=f(b)-f(a) $$ Thus, we have shown that the flux form of Green's Theorem applied to the given vector field results in the Fundamental Theorem of Calculus.

Key concepts

Fundamental Theorem of Calculus

When embarking on the adventure of calculus, the Fundamental Theorem of Calculus (FTC) stands as a monumental landmark. It paints a splendid picture of the intrinsic relationship between differentiation and integration. To state it succinctly, FTC claims that if you have a continuous function, let's call it f(x), and you take its integral from a to b, what you're doing is essentially unwrapping the accumulated value that f has collected as x travels from a to b.

Mathematically, the theorem tells us that:
\[ \int_{a}^{b} f'(x) \, dx = f(b) - f(a) \]

Imagine it as setting out on a journey at mile marker a (this is f(a)), hiking up to mile marker b (this is f(b)), and the integral—well, that's the total ground covered. In the context of our exercise, the function's derivative, f'(x), is integrated, and we end up with the net change in function values over the interval [a, b].

Flux form of Green's Theorem

Picture a flowing river—a vector field is much like that, but with flow happening in all sorts of directions and with varying intensity. Now, if you want to scoop out a certain amount of water as it flows past a loop, that's where the Flux form of Green's Theorem comes into play, tactfully extending a helping hand.

Think of it as a superpower that tells you how much the vector field—the flow of our conceptual river—is sweeping across a curve. The theorem asserts that the circulation along a closed curve C, which wraps around a region D, can be converted into a double integral over the region's area. The flux form is eloquently expressed as:
\[ \oint_C \textbf{F} \cdot \textbf{n} \, ds = \iint_D (abla \times \textbf{F}) \, dA \]

This means you can switch between calculating a pesky line integral along the loop, C, to calculating the curl of the vector field over the region it ensnares, D. In our exercise, we witnessed this theorem simplifying the task by showing that the curl of our particular field was zero, and, voila, the seemingly complicated became straightforward!

Curl of a vector field

Spin around with your arms out—that whirling motion is akin to the curl in a vector field. Except, instead of arms, it's more about how the field twists and turns in space. Curl measures this tendency of rotation at any given point.

To catch the curl of a vector field G, given by G(x, y) = \( \langle P(x,y), Q(x,y) \rangle \), you need to perform a bit of calculus magic, specifically a cross-derivative dance. Here's the spell: \[ abla \times \textbf{G} = \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \]

When this difference equals zero, it means we've hit a curl-less field, much like a calm lake with no whirlpools. In the problem we delved into, it was precisely this absence of curl (since P and Q were partial derivatives of constants) that led us to the tranquil simplicity of the Fundamental Theorem of Calculus.

Line integral

Imagine taking a stroll along a path and tallying up values you encounter—let's say, points for every flower spotted. This is essentially what a line integral does; it's an accumulation of function values along a specified curve.

Here, you're not just counting or adding up—a line integral intricately weaves together the function values with the path's geometry, resulting in a total sum that reflects the landscape of the function over that journey. In a neat formula, the line integral for a vector field F along a curve C is:
\[ \oint_C \textbf{F} \cdot d\textbf{r} \]

In our foray through Green's Theorem and the rectangular path around region R, we split this path into digestible segments, calculating each one's line integral separately. Though some of the segments contributed nothing (zero), one segment did the heavy lifting, leading us to the concluding epiphany of the Fundamental Theorem connection.