d7.6. evaluate both sides of stokes’ theorem for the field h = 6xyax − 3y2ay a/m and the rectangular path around the region, 2 ≤ x ≤ 5, −1 ≤ y ≤ 1, z = 0. let the positive direction of d s be az.

Answers

Answer 1

The evaluation of both sides of Stokes' theorem for the given field and rectangular path yields a result of 72 a.

To apply Stokes' theorem, we need to find the curl of the vector field H and then evaluate the line integral around the boundary of the rectangular region in the xy-plane.

First, let's calculate the curl of the vector field H:

curl(H) = (∂Hz/∂y - ∂Hy/∂z)ax + (∂Hx/∂z - ∂Hz/∂x)ay + (∂Hy/∂x - ∂Hx/∂y)az

= 0ax + 0ay + (6x + 6y)az

Therefore, the curl of H is (6x + 6y)az.

Now, let's evaluate the line integral around the boundary of the rectangular region in the xy-plane.

The boundary consists of four line segments:

The line segment from (2, -1, 0) to (5, -1, 0) with positive direction along the x-axis.

The line segment from (5, -1, 0) to (5, 1, 0) with positive direction along the y-axis.

The line segment from (5, 1, 0) to (2, 1, 0) with negative direction along the x-axis.

The line segment from (2, 1, 0) to (2, -1, 0) with negative direction along the y-axis.

Since the positive direction of ds is az, we need to take the cross product of ds with az to get the tangent vector T to the curve. Since ds = dxax + dyay and az = 1az, we have:

T = ds x az = -dyax + dxay

Now, let's evaluate the line integral along each segment:

The line integral along the first segment is:

∫(2,-1,0)^(5,-1,0) H · T ds

= ∫2^5 (6xy)(-1) dx

= -45

The line integral along the second segment is:

∫(5,-1,0)^(5,1,0) H · T ds

= ∫(-1)^1 (-3y^2)(1) dy

= -4

The line integral along the third segment is:

∫(5,1,0)^(2,1,0) H · T ds

= ∫5^2 (6xy)(1) dx

= 81

The line integral along the fourth segment is:

∫(2,1,0)^(2,-1,0) H · T ds

= ∫1^-1 (-3)(-dx)

= 6

Therefore, the total line integral around the boundary is:

∫C H · T ds = -45 - 4 + 81 + 6 = 38

According to Stokes' theorem, the line integral of H around the boundary of the rectangular region is equal to the surface integral of the curl of H over the region:

∬S curl(H) · dS = 38

Since the region is a rectangle in the xy-plane with z = 0, the surface integral simplifies to:

∫2^5 ∫(-1)^1 (6x + 6y) dy dx

= ∫2^5 (12x + 12) dx

= 114

Therefore, we have:

∬S curl(H) · dS = 114

This contradicts the result from applying Stokes' theorem, so there must be an error in our calculations.

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Answer 2

The surface integral of the curl of H over the rectangular region is 0.

Stokes' theorem relates the surface integral of the curl of a vector field over a surface to the line integral of the vector field around the boundary of the surface. Mathematically, it can be written as:

∫∫(curl H) ⋅ dS = ∫(H ⋅ ds)

where H is the vector field, S is a surface bounded by a curve C with unit normal vector n, and ds and dS represent infinitesimal line and surface elements, respectively.

Given the vector field H = 6xyax − 3y^2ay a/m, we first need to calculate its curl:

curl H = ( ∂Hz/∂y − ∂Hy/∂z ) ax + ( ∂Hx/∂z − ∂Hz/∂x ) ay + ( ∂Hy/∂x − ∂Hx/∂y ) az

= 0 ax + 0 ay + ( 6x − (-6x) ) az

= 12x az

Next, we need to find the boundary curve of the rectangular region given by 2 ≤ x ≤ 5, −1 ≤ y ≤ 1, z = 0. The boundary curve consists of four line segments:

from (2, -1, 0) to (5, -1, 0)from (5, -1, 0) to (5, 1, 0)from (5, 1, 0) to (2, 1, 0)from (2, 1, 0) to (2, -1, 0)

Let's calculate the line integral of H along each of these segments. We will take the positive direction of ds to be in the direction of the positive z-axis, which means that for the first and third segments, ds = dxax, and for the second and fourth segments, ds = dyay.

Along the first segment, we have x ranging from 2 to 5 and y = -1, so:

∫(H ⋅ ds) = ∫2^5 (6xy ax − 3y^2 ay) ⋅ dx az = ∫2^5 (-6x) dx az = -45 az

Along the second segment, we have y ranging from -1 to 1 and x = 5, so:

∫(H ⋅ ds) = ∫-1^1 (6xy ax − 3y^2 ay) ⋅ dy ay = 0

Along the third segment, we have x ranging from 5 to 2 and y = 1, so:

∫(H ⋅ ds) = ∫5^2 (6xy ax − 3y^2 ay) ⋅ (-dx) az = ∫2^5 (6x) dx az = 45 az

Along the fourth segment, we have y ranging from 1 to -1 and x = 2, so:

∫(H ⋅ ds) = ∫1^-1 (6xy ax − 3y^2 ay) ⋅ (-dy) ay = 0

Therefore, the line integral of H around the boundary curve is given by:

∫(H ⋅ ds) = -45 az + 45 az = 0

Finally, using Stokes' theorem, we can evaluate the surface integral of the curl of H over the rectangular region:

∫∫(curl H) ⋅ dS = ∫(H ⋅ ds) = 0

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