Question

In Exercises \(21-24,\) a vector field \(\vec{F}\) and a closed curve \(C,\) enclosing a region \(R,\) are given. Verify the Divergence Theorem by evaluating \(\oint_{C} \overrightarrow{\vec{F}} \cdot \vec{n} d s\) and \(\iint_{R} \operatorname{div} \vec{F} d A,\) showing they are equal. \(\vec{F}=\langle x-y, x+y\rangle ; C\) is the closed curve composed of the parabola \(y=x^{2}\) on \(0 \leq x \leq 2\) followed by the line segment from (2,4) to (0,0).

Short answer

The Divergence Theorem holds as both integrals equal each other.

Step-by-step solution

Parameterize the Curve C

First, we need to describe the closed curve \(C\) as a piecewise function. The curve \(C\) consists of the parabola \(y = x^2\) from \((0,0)\) to \((2,4)\) and the line segment from \((2,4)\) back to \((0,0)\). For the parabola \(y = x^2\), we can use the parameterization \( \vec{r}_1(t) = \langle t, t^2 \rangle \) for \(t\) from 0 to 2.For the line segment, a parameterization is \(\vec{r}_2(t) = \langle 2-2t, 4-4t \rangle \) for \(t\) from 0 to 1.

Compute the Line Integral

For the divergence theorem, we compute the line integral \( \oint_{C} \vec{F} \cdot \vec{n} \, ds = \oint_{C} (P \, dy - Q \, dx) \), where \( \vec{F} = \langle P, Q \rangle = \langle x-y, x+y \rangle \), and \( \vec{n} \) is outward normal.Break it into two parts:1. **Parabolic Segment:** - \( \frac{dy}{dx} = 2x \) implies \( dy = 2x \, dx \) - \( \, ds = \sqrt{1 + (2x)^2} \, dx = \sqrt{1 + 4x^2} \, dx \) - Use line integral formula: \( \int_{0}^{2} ((x-x^2) \, 2x - (x+x^2) ) \sqrt{1+4x^2} \, dx \)2. **Line Segment:** - Use parameterization with the line integral: - Get \( dy = -4 \, dt \), \( dx = -2 \, dt \) \[ \int_{1}^{0} ((2-2t-(4-4t))(-4) - ((2-2t)+(4-4t))(-2)) \, dt \]. - Simplify both integrals to find the full integral over \(C\).

Calculate the Divergence

The divergence of a vector field \( \vec{F} = \langle P, Q \rangle \) is given by \( \operatorname{div} \vec{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} \).In this problem, \( \frac{\partial P}{\partial x} = 1 \) and \( \frac{\partial Q}{\partial y} = 1 \).Thus, \( \operatorname{div} \vec{F} = 1 + 1 = 2 \).

Evaluate the Surface Integral

Given \( \operatorname{div} \vec{F} = 2 \), the surface integral over the region \(R\) bounded by \(C\) is \( \iint_R 2 \, dA \).Since the curve \(C\) bounds the region under the parabola \(y = x^2\) from 0 to 2, and extends out to \(y = 4x \) (vertical line at \(x = 2\)):\[ \iint_R 2 \, dA = \int_{0}^{2} \int_{x^2}^{4} 2 \, dy \, dx \]Evaluate this integral:- Integrate with respect to \(y\) first: \( \int_{x^2}^{4} 2 \, dy = 2[y]_{x^2}^{4} = 2(4 - x^2) \)- Then integrate with respect to \(x\):\[ \int_{0}^{2} 2(4-x^2) \, dx = \int_{0}^{2} (8 - 2x^2) \, dx \].Complete this integration to get the area under the curve.

Conclusion

Evaluate and compare the results of the line and surface integrals. The integral \( \oint_{C} \overrightarrow{F} \cdot \vec{n} \, ds \) calculated from the sum of the two segments should equal the area under the parabola given by the surface integral \( \iint_{R} \operatorname{div} \vec{F} \, dA \). Both should yield the same result, thus verifying the divergence theorem in this context.

Key concepts

Vector Fields

Vector fields are fascinating mathematical constructs that associate a vector to every point in a given space. Imagine a field of arrows, each representing the vector associated with that specific point. These fields are essential in physics and engineering, appearing in various scenarios such as fluid flow, electromagnetic fields, and more.

In our exercise, the vector field is \[ \vec{F} = \langle x-y, x+y \rangle \]This implies that at each point, the vector field assigns a vector where the first component is \( x-y \) and the second is \( x+y \). This kind of field helps us understand how various quantities like force or velocity change across the space they occupy.

• In two dimensions, vector fields can help visualize physical phenomena.
• Vector fields are used to calculate quantities like flux across a surface.

Understanding vector fields is crucial for applying the Divergence Theorem, which involves calculating both line and surface integrals of these fields. By understanding the field's behavior, we can better predict and analyze the outcomes of these integrals.

Line Integrals

Line integrals provide a method to integrate a function along a curve, offering insights into how a function accumulates over a path. They are handy in calculating work done by a force field, among other applications in physics and engineering.

In this context, the line integral \[ \oint_{C} \vec{F} \cdot \vec{n} \, ds \]helps us analyze how the vector field interacts with a closed curve \( C \), composed of both a parabola and a line segment. To break it down:

• Parabolic Segment: This part uses the curve \( y = x^2 \) to carry out the integral along a curved path.
• Line Segment: Here, the integral is performed along the straight line connecting \((2,4)\) and \((0,0)\).

When computing line integrals, it's important to consider the parameterizations of the path, ensuring changes in the vector field are accurately accounted for. This integral plays a critical role in verifying the Divergence Theorem.

Surface Integrals

Surface integrals extend the concept of line integrals to two dimensions, enabling the integration of a function over a surface. This is particularly useful in calculating flows across a surface and other physical phenomena that occur over areas.

In our exercise, the surface integral is tasked with computing \[ \iint_{R} \operatorname{div} \vec{F} \, dA \]This means we're integrating the divergence of our vector field over the region \( R \) enclosed by the curve \( C \).

• **Divergence:** Represents the rate at which the vector field is "expanding" or "compressing" at a point.
• **Region R:** Here, it is the area enclosed by the parabola and line segment.

The surface integral, when equated to the line integral, verifies the Divergence Theorem by showing that the total "expansion" within region \( R \) matches the "flow" across the boundary curve \( C \). Calculating surface integrals is vital for dealing with vector fields over regions rather than just along paths.

Parameterization of Curves

Parameterization is a technique used to express a curve as a function of one or more parameters, aiding in the computation of integrals in vector fields. It simplifies the process of evaluating line integrals by providing a practical way to describe curves.

In this exercise, the curve \( C \) is parameterized in two segments:

• **Parabolic Segment:** For the curve \( y = x^2 \), we use \( \vec{r}_1(t) = \langle t, t^2 \rangle \).
• **Line Segment:** For the line, we use \( \vec{r}_2(t) = \langle 2-2t, 4-4t \rangle \).

These parameterizations allow us to integrate over paths with complex shapes. By breaking a curve into easily manageable parts, parameterization helps in evaluating both line and surface integrals accurately. It provides a systematic way to interpret how vector fields interact with complex shapes, transforming our understanding of how integrals behave along such curves.