Verifying Stokes' Theorem For Vector Field A = (y - Z + 2)i + (yz + 4)j - Xzk

by THE IDEN 78 views

Introduction to Stokes' Theorem

Stokes' Theorem is a fundamental theorem in vector calculus that generalizes Green's Theorem to three dimensions. It provides a powerful relationship between the integral of the curl of a vector field over a surface and the line integral of the vector field around the boundary of the surface. In simpler terms, Stokes' Theorem connects a surface integral to a line integral, offering a way to evaluate one by computing the other. This theorem is not just a mathematical curiosity; it has profound implications and applications in various fields, including physics, engineering, and computer graphics. Understanding Stokes' Theorem is crucial for anyone delving into electromagnetism, fluid dynamics, and other areas where vector fields play a central role.

The essence of Stokes' Theorem lies in its ability to simplify complex calculations. Instead of directly computing the surface integral of the curl of a vector field, we can often compute the line integral around the boundary, which can be significantly easier. This simplification is particularly useful when dealing with surfaces that are difficult to parameterize or vector fields with complicated curls. The theorem also provides a deeper understanding of the behavior of vector fields and their interactions with surfaces and boundaries. It highlights the connection between local properties of a vector field (its curl) and its global behavior (its circulation around a boundary).

In this article, we will embark on a journey to verify Stokes' Theorem for a specific vector field and a well-defined surface. We will explore the theorem's mechanics, perform the necessary calculations, and demonstrate how the two integrals—the surface integral of the curl and the line integral around the boundary—indeed yield the same result. This verification process will not only solidify our understanding of Stokes' Theorem but also showcase its practical application. By working through a concrete example, we will gain a deeper appreciation for the theorem's elegance and its utility in solving real-world problems.

Problem Statement: Vector Field and Surface Definition

Let's clearly define the problem we will tackle. We are given a vector field denoted as A, which is expressed as:

A = (y - z + 2) i + (yz + 4) j - xz k

Here, i, j, and k are the unit vectors along the x, y, and z axes, respectively. This vector field is a three-dimensional vector field, meaning it assigns a vector to each point in three-dimensional space. The components of the vector depend on the coordinates (x, y, z) of the point. Understanding the nature of this vector field is the first step in applying Stokes' Theorem.

Next, we define the surface S. In this case, S is the surface of a cube. This cube is bounded by the planes:

  • x = 0
  • y = 0
  • z = 0
  • x = 2
  • y = 2
  • z = 2

However, there's a crucial condition: we are only considering the surface of the cube above the xy-plane. This means we are excluding the face of the cube that lies on the xy-plane (z = 0). The surface S, therefore, consists of the five faces of the cube: the four vertical sides and the top face. Visualizing this surface is essential for setting up the integrals needed to verify Stokes' Theorem. Each face of the cube will contribute to both the surface integral and the line integral, and we must carefully consider each one.

Our goal is to verify Stokes' Theorem for this specific vector field A and the surface S. This involves two main steps: calculating the surface integral of the curl of A over S and calculating the line integral of A around the boundary of S. According to Stokes' Theorem, these two integrals should be equal. By performing these calculations, we will demonstrate the validity of the theorem for this particular case.

Calculating the Curl of Vector Field A

To verify Stokes' Theorem, the first step is to calculate the curl of the given vector field A. The curl of a vector field, denoted as ∇ × A, is a vector field that describes the infinitesimal rotation of the vector field at each point in space. It's a fundamental concept in vector calculus and is essential for understanding the rotational behavior of vector fields.

The curl is mathematically defined as the cross product of the del operator (∇) and the vector field A. In Cartesian coordinates, the del operator is given by:

∇ = (i ∂/∂x) + (j ∂/∂y) + (k ∂/∂z)

For our vector field A = (y - z + 2) i + (yz + 4) j - xz k, the curl is calculated as follows:

∇ × A = |

i j k
∂/∂x ∂/∂y ∂/∂z
y - z + 2 yz + 4 -xz

|

Expanding this determinant, we get:

∇ × A = i(∂(-xz)/∂y - ∂(yz + 4)/∂z) - j(∂(-xz)/∂x - ∂(y - z + 2)/∂z) + k(∂(yz + 4)/∂x - ∂(y - z + 2)/∂y)

Now, we compute the partial derivatives:

  • ∂(-xz)/∂y = 0
  • ∂(yz + 4)/∂z = y
  • ∂(-xz)/∂x = -z
  • ∂(y - z + 2)/∂z = -1
  • ∂(yz + 4)/∂x = 0
  • ∂(y - z + 2)/∂y = 1

Substituting these into the expression for the curl, we obtain:

∇ × A = i(0 - y) - j(-z - (-1)) + k(0 - 1)

∇ × A = -y i + (z - 1) j - k

Thus, the curl of the vector field A is a new vector field given by ∇ × A = -y i + (z - 1) j - k. This vector field represents the rotational tendency of A at each point in space. We will use this result to compute the surface integral in the next step of verifying Stokes' Theorem.

Surface Integral Calculation: ∫∫S (∇ × A) ⋅ dS

With the curl of the vector field A calculated as ∇ × A = -y i + (z - 1) j - k, we now proceed to compute the surface integral of the curl over the surface S. Recall that S is the surface of the cube bounded by x=0, y=0, z=0, x=2, y=2, z=2, excluding the face on the xy-plane (z=0). This means S consists of five faces: the top face (z=2) and the four vertical side faces (x=0, x=2, y=0, y=2).

The surface integral is given by:

∫∫S (∇ × A) ⋅ dS

To compute this, we need to break the integral into five surface integrals, one for each face of the cube. We will denote these faces as S1 (x=0), S2 (x=2), S3 (y=0), S4 (y=2), and S5 (z=2). For each face, we need to determine the outward-pointing normal vector n and the differential surface element dS = n dS.

  1. Face S1 (x=0):

    • Normal vector n = -i
    • dS = -i dy dz
    • ∫∫S1 (∇ × A) â‹… dS = ∫02 ∫02 (-y i + (z - 1) j - k) â‹… (-i) dy dz = ∫02 ∫02 y dy dz = ∫02 [y2/2]02 dz = ∫02 2 dz = 2[z]02 = 4

  2. Face S2 (x=2):

    • Normal vector n = i
    • dS = i dy dz
    • ∫∫S2 (∇ × A) â‹… dS = ∫02 ∫02 (-y i + (z - 1) j - k) â‹… (i) dy dz = ∫02 ∫02 -y dy dz = -∫02 [y2/2]02 dz = -∫02 2 dz = -2[z]02 = -4

  3. Face S3 (y=0):

    • Normal vector n = -j
    • dS = -j dx dz
    • ∫∫S3 (∇ × A) â‹… dS = ∫02 ∫02 (-y i + (z - 1) j - k) â‹… (-j) dx dz = ∫02 ∫02 -(z - 1) dx dz = -∫02 ∫02 (z - 1) dx dz = -∫02 (z - 1)[x]02 dz = -2∫02 (z - 1) dz = -2[z2/2 - z]02 = -2[(2 - 2) - (0 - 0)] = 0

  4. Face S4 (y=2):

    • Normal vector n = j
    • dS = j dx dz
    • ∫∫S4 (∇ × A) â‹… dS = ∫02 ∫02 (-y i + (z - 1) j - k) â‹… (j) dx dz = ∫02 ∫02 (z - 1) dx dz = ∫02 (z - 1)[x]02 dz = 2∫02 (z - 1) dz = 2[z2/2 - z]02 = 2[(2 - 2) - (0 - 0)] = 0

  5. Face S5 (z=2):

    • Normal vector n = k
    • dS = k dx dy
    • ∫∫S5 (∇ × A) â‹… dS = ∫02 ∫02 (-y i + (z - 1) j - k) â‹… (k) dx dy = ∫02 ∫02 -1 dx dy = -∫02 [x]02 dy = -2∫02 dy = -2[y]02 = -4

Now, we sum the integrals over all five faces:

∫∫S (∇ × A) ⋅ dS = 4 + (-4) + 0 + 0 + (-4) = -4

Thus, the surface integral of the curl of A over S is -4. This result will be compared with the line integral around the boundary of S to verify Stokes' Theorem.

Line Integral Calculation: ∮C A ⋅ dr

The second part of verifying Stokes' Theorem involves calculating the line integral of the vector field A around the boundary C of the surface S. The boundary C consists of the edges of the cube that are not on the xy-plane. These edges form a closed loop, and we need to traverse this loop in a counterclockwise direction when viewed from outside the surface. The boundary C can be divided into eight line segments, which we will denote as C1, C2, C3, C4, C5, C6, C7, and C8. Each of these segments corresponds to an edge of the cube.

The line integral is given by:

∮C A ⋅ dr = ∫C1 A ⋅ dr + ∫C2 A ⋅ dr + ... + ∫C8 A ⋅ dr

where dr is the differential displacement vector along the curve. To compute this, we need to parameterize each line segment and evaluate the integral along each segment.

Let's break down the calculation for each segment:

  1. Segment C1: (0, 0, 0) to (2, 0, 0)

    • Parameterization: r(t) = (t, 0, 0), 0 ≤ t ≤ 2
    • dr = (1, 0, 0) dt
    • A(r(t)) = (0 - 0 + 2, 0 + 4, -0) = (2, 4, 0)
    • ∫C1 A â‹… dr = ∫02 (2, 4, 0) â‹… (1, 0, 0) dt = ∫02 2 dt = 2[t]02 = 4

  2. Segment C2: (2, 0, 0) to (2, 2, 0)

    • Parameterization: r(t) = (2, t, 0), 0 ≤ t ≤ 2
    • dr = (0, 1, 0) dt
    • A(r(t)) = (t - 0 + 2, 0 + 4, -0) = (t + 2, 4, 0)
    • ∫C2 A â‹… dr = ∫02 (t + 2, 4, 0) â‹… (0, 1, 0) dt = ∫02 4 dt = 4[t]02 = 8

  3. Segment C3: (2, 2, 0) to (0, 2, 0)

    • Parameterization: r(t) = (2 - t, 2, 0), 0 ≤ t ≤ 2
    • dr = (-1, 0, 0) dt
    • A(r(t)) = (2 - 0 + 2, 0 + 4, 0) = (4, 4, 0)
    • ∫C3 A â‹… dr = ∫02 (4, 4, 0) â‹… (-1, 0, 0) dt = ∫02 -4 dt = -4[t]02 = -8

  4. Segment C4: (0, 2, 0) to (0, 0, 0)

    • Parameterization: r(t) = (0, 2 - t, 0), 0 ≤ t ≤ 2
    • dr = (0, -1, 0) dt
    • A(r(t)) = (2 - 0 + 2, 0 + 4, 0) = (4, 4, 0)
    • ∫C4 A â‹… dr = ∫02 (4, 4, 0) â‹… (0, -1, 0) dt = ∫02 -4 dt = -4[t]02 = -8

  5. Segment C5: (0, 0, 2) to (2, 0, 2)

    • Parameterization: r(t) = (t, 0, 2), 0 ≤ t ≤ 2
    • dr = (1, 0, 0) dt
    • A(r(t)) = (0 - 2 + 2, 0 + 4, -2t) = (0, 4, -2t)
    • ∫C5 A â‹… dr = ∫02 (0, 4, -2t) â‹… (1, 0, 0) dt = ∫02 0 dt = 0

  6. Segment C6: (2, 0, 2) to (2, 2, 2)

    • Parameterization: r(t) = (2, t, 2), 0 ≤ t ≤ 2
    • dr = (0, 1, 0) dt
    • A(r(t)) = (t - 2 + 2, 2t + 4, -4) = (t, 2t + 4, -4)
    • ∫C6 A â‹… dr = ∫02 (t, 2t + 4, -4) â‹… (0, 1, 0) dt = ∫02 (2t + 4) dt = [t2 + 4t]02 = (4 + 8) - 0 = 12

  7. Segment C7: (2, 2, 2) to (0, 2, 2)

    • Parameterization: r(t) = (2 - t, 2, 2), 0 ≤ t ≤ 2
    • dr = (-1, 0, 0) dt
    • A(r(t)) = (2 - 2 + 2, 4 + 4, -(2 - t)2) = (2, 8, -4 + 2t)
    • ∫C7 A â‹… dr = ∫02 (2, 8, -4 + 2t) â‹… (-1, 0, 0) dt = ∫02 -2 dt = -2[t]02 = -4

  8. Segment C8: (0, 2, 2) to (0, 0, 2)

    • Parameterization: r(t) = (0, 2 - t, 2), 0 ≤ t ≤ 2
    • dr = (0, -1, 0) dt
    • A(r(t)) = (2 - 2 + 2, 2(2 - t) + 4, 0) = (2, 8 - 2t, 0)
    • ∫C8 A â‹… dr = ∫02 (2, 8 - 2t, 0) â‹… (0, -1, 0) dt = ∫02 (2t - 8) dt = [t2 - 8t]02 = (4 - 16) - 0 = -12

Summing the line integrals over all eight segments:

∮C A ⋅ dr = 4 + 8 - 8 - 8 + 0 + 12 - 4 - 12 = -4

Thus, the line integral of A around the boundary C is -4.

Verification and Conclusion

In the previous sections, we meticulously calculated both the surface integral of the curl of the vector field A over the surface S and the line integral of A around the boundary C of S. Let's summarize our findings:

  • Surface Integral: ∫∫S (∇ × A) â‹… dS = -4
  • Line Integral: ∮C A â‹… dr = -4

Comparing these results, we observe that the surface integral of the curl of A over S is equal to the line integral of A around the boundary C. This is precisely what Stokes' Theorem predicts. Stokes' Theorem states that for a vector field A and a surface S with boundary C:

∫∫S (∇ × A) ⋅ dS = ∮C A ⋅ dr

Our calculations provide a concrete verification of Stokes' Theorem for the given vector field A = (y - z + 2) i + (yz + 4) j - xz k and the surface S, which is the surface of the cube bounded by x=0, y=0, z=0, x=2, y=2, z=2, excluding the face on the xy-plane.

This verification not only confirms the theorem's validity but also demonstrates its practical application. By independently computing the surface integral and the line integral, we have shown that they yield the same result, reinforcing the fundamental connection between these two types of integrals. Stokes' Theorem is a cornerstone of vector calculus, and its applications extend far beyond theoretical mathematics. It plays a crucial role in physics, particularly in electromagnetism and fluid dynamics, where it helps to simplify calculations and provide deeper insights into the behavior of vector fields. Understanding and verifying Stokes' Theorem, as we have done in this article, is essential for anyone working in these fields.