Problem 16 - Advanced Engineering Mathematics

Question

Use Stokes' theorem to evaluate \(\iint_{S}(\operatorname{curl} \mathbf{F}) \cdot \mathbf{n} d S\). Assume that the surface \(S\) is oriented upward. \(\mathbf{F}=2 x y^{2} z \mathbf{i}+2 x^{2} y z \mathbf{j}+\left(x^{2} y^{2}-6 x\right) \mathbf{k} ; S\) that portion of the plane \(z=y\) that lies inside the cylinder \(x^{2}+y^{2}=1\)

Step-by-Step Solution

Answer

The integral evaluates to 0.

Step 1: Understand Stokes' Theorem

Stokes' Theorem states that \( \iint_{S} (abla \times \mathbf{F}) \cdot \mathbf{n} \, dS = \oint_{C} \mathbf{F} \cdot d\mathbf{r} \), where \( C \) is the curve bounding the surface \( S \). You convert the surface integral into a line integral over the boundary \( C \).

Step 2: Identify the Boundary Curve

The plane \( z = y \) within the cylinder \( x^2 + y^2 = 1 \) is bounded by the circle defined by \( x^2 + y^2 = 1 \). This circle is the boundary curve \( C \) on the surface \( z = y \).

Step 3: Parameterize the Boundary Curve

The boundary curve \( C \) is a circle, so parameterize it with \( \mathbf{r}(t) = \langle \cos t, \sin t, \sin t \rangle \) for \( t \) in \( [0, 2\pi] \). This is because \( z = y \) on \( C \).

Step 4: Compute \( \nabla \times \mathbf{F} \)

Calculate the curl of \( \mathbf{F} \): \( abla \times \mathbf{F} = \left( \frac{\partial}{\partial y}(x^2y^2 - 6x) - \frac{\partial}{\partial z}(2x^2yz), \frac{\partial}{\partial z}(2xy^2z) - \frac{\partial}{\partial x}(x^2y^2 - 6x), \frac{\partial}{\partial x}(2x^2yz) - \frac{\partial}{\partial y}(2xy^2z) \right) \).

Step 5: Evaluate the Curl

Simplify the expression for each component from Step 4:- \( \frac{\partial}{\partial y}(x^2y^2 - 6x) = 2x^2y \), \( \frac{\partial}{\partial z}(2x^2yz) = 2x^2y \), so the first component is \( 2x^2y - 2x^2y = 0 \).- \( \frac{\partial}{\partial z}(2xy^2z) = 2xy^2 \), \( \frac{\partial}{\partial x}(x^2y^2 - 6x) = 2xy^2 - 6 \), so the second component is \( 2xy^2 - 2xy^2 + 6 = 6 \).- \( \frac{\partial}{\partial x}(2x^2yz) = 4xyz \), \( \frac{\partial}{\partial y}(2xy^2z) = 4xyz \), so the third component is \( 4xyz - 4xyz = 0 \).Thus, \( abla \times \mathbf{F} = \langle 0, 6, 0 \rangle \).

Step 6: Compute Line Integral

With \( abla \times \mathbf{F} = \langle 0, 6, 0 \rangle \), the line integral becomes \( \oint_{C} \mathbf{F} \cdot d\mathbf{r} \). Calculate \( d\mathbf{r} = \langle -\sin t, \cos t, \cos t \rangle dt \). Since \( \mathbf{F} \) in line integral reduces to 6 in the direction of \( \mathbf{j} \), the integral becomes: \( \oint_{C} 6 \cos t \, dt \).

Step 7: Solve the Line Integral

Solve the integral: \( \oint_{C} 6 \cos t \, dt = 6 \int_0^{2\pi} \cos t \, dt = 6 \left[ \sin t \right]_0^{2\pi} = 6 (0 - 0) = 0 \).

Step 8: Conclude with Stokes' Theorem

Based on Stokes' theorem and calculating \( \oint_{C} \mathbf{F} \cdot d\mathbf{r} = 0 \), we conclude that \( \iint_{S} (abla \times \mathbf{F}) \cdot \mathbf{n} \, dS = 0 \).

Key Concepts

Surface Integral

In vector calculus, the concept of a surface integral extends the idea of an integral to functions that are defined over a surface. Surface integrals help to compute the flow (or flux) of a vector field through a given surface. The vector function is integrated over the surface, taking into account the angle between the vector field and the surface's normal vector at each point.

For a surface integral of a vector field, denoted as \( \iint_{S} \mathbf{F} \cdot \mathbf{n} \, dS \), where \( \mathbf{F} \) is the vector field, \( \mathbf{n} \) represents the unit normal vector to the surface \( S \), and \( dS \) is an infinitesimally small area element of the surface.

• It is useful in physics for calculating quantities like electric flux or magnetic flux over a surface.
• Surface integrals consider both the magnitude of \( \mathbf{F} \) and its orientation relative to \( \mathbf{n} \).

Stokes’ Theorem often simplifies the computation of surface integrals by converting them into line integrals.

Curl of a Vector Field

The curl of a vector field is a measure of the field's tendency to rotate about a point—a key concept in vector calculus. Mathematically, the curl of a vector field \( \mathbf{F} \) is another vector field and is denoted as \( abla \times \mathbf{F} \). The output value at any given point gives the axis of rotation and the magnitude of curling at that point.

To compute the curl for a vector field \( \mathbf{F} = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k} \), use the formula:\[abla \times \mathbf{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right) \mathbf{i} + \left( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} \right) \mathbf{j} + \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \mathbf{k}.\]

• It is extensively used in electromagnetism and fluid dynamics to describe rotational fields.
• The curl itself can be viewed as the circulation density at each point.

In the context of Stokes' Theorem, the curl is integrated over a surface to relate it to a line integral along the boundary of the surface.

Boundary Curve Parameterization

When evaluating integrals over a surface, knowing how to parameterize the boundary curve is crucial. A boundary curve \( C \) is the closed loop surrounding the surface, and parameterization involves expressing points on the curve as a function of a single parameter \( t \). This essentially traces the loop.

In the given exercise, the boundary curve is a circle on the plane \( z = y \), with the equation \( x^2 + y^2 = 1 \). Therefore, the curve can be parameterized as \( \mathbf{r}(t) = \langle \cos t, \sin t, \sin t \rangle \) with \( t \) ranging from 0 to \( 2\pi \).

• Parameterization helps in expressing line integrals as standard integrals over a single parameter.
• It effectively transforms the line integral computation into evaluating the integral of a function of \( t \) over a defined interval.

Proper parameterization ensures that all points on the curve are accounted for and makes computations tractable.

Line Integral

A line integral computes the sum of field values along a curve, taking into account both the field and the path's direction. In vector calculus, line integrals are used to evaluate integrals over paths in a vector field, denoted as \( \oint_{C} \mathbf{F} \cdot d\mathbf{r} \), where \( \mathbf{F} \) is the vector field and \( d\mathbf{r} \) is an infinitesimal displacement vector along the curve \( C \).

The value of the line integral depends on:

• The path \( C \), including its direction.
• The vector field \( \mathbf{F} \) and its components.

Line integrals appear in physics when calculating work done by a force field, with the force represented by \( \mathbf{F} \) and \( d\mathbf{r} \) as the differential motion in the field's direction.

When using Stokes' Theorem, the line integral becomes the key to solving the surface integral by evaluating the curl of \( \mathbf{F} \) over the loop enclosing the surface.