Problem 16 - Advanced Engineering Mathematics

Question

In Problems 13-16, use Stokes' theorem to evaluate \(\iint_{S}\) (curl \(\left.\mathbf{F}\right) \cdot \mathbf{n} d S\). 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 using Stokes' theorem and symmetry.

Step 1: Understanding Stokes' Theorem

Stokes' Theorem relates a surface integral of the curl of a vector field over a surface \(S\) to a line integral of the vector field over the boundary \(C\) of \(S\). It states that \( \iint_S \text{curl } \mathbf{F} \cdot d\mathbf{S} = \oint_C \mathbf{F} \cdot d\mathbf{r} \). Our task is to evaluate the surface integral using this theorem.

Step 2: Calculating the Curl of \( \mathbf{F} \)

Calculate \( \text{curl } \mathbf{F} = abla \times \mathbf{F} \). Given \( \mathbf{F} = 2xy^2z \mathbf{i} + 2x^2yz \mathbf{j} + (x^2y^2 - 6x) \mathbf{k} \), we use the determinant formula:\[ \text{curl } \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ 2xy^2z & 2x^2yz & x^2y^2 - 6x \end{vmatrix} \]Calculate each component using the determinant.

Step 3: Setting Up the Line Integral

Identify the orientation and parametrization for the boundary \(C\) of the surface \(S\). Since \(S\) is the portion of the plane \(z = y\) inside the cylinder \(x^2 + y^2 = 1\), its boundary \(C\) is the circle \(x^2 + y^2 = 1\). Parametrize this using \( \mathbf{r}(t) = \cos t \, \mathbf{i} + \sin t \, \mathbf{j} + \sin t \, \mathbf{k} \) for \(t\) from \(0\) to \(2\pi\).

Step 4: Evaluating the Line Integral

Substitute the parametrization \( \mathbf{r}(t) \) into \( \mathbf{F} \) and evaluate \( \oint_C \mathbf{F} \cdot d\mathbf{r} \). Compute \( d\mathbf{r} = \left(-\sin t \, \mathbf{i} + \cos t \, \mathbf{j} + \cos t \, \mathbf{k}\right) dt \) and find \( \mathbf{F} \cdot d\mathbf{r} \). Integrate over \(t\) from 0 to \(2\pi\).

Step 5: Solving the Integral

Perform the integration of \( \mathbf{F} \cdot d\mathbf{r} \) over the interval \([0, 2\pi]\). Due to symmetry and trigonometric properties, simplify the expressions to evaluate the integral.

Key Concepts

Curl of a Vector Field

The curl of a vector field is a very important concept. It measures the rotational tendency of the field at a point. Imagine the vector field as a series of arrows representing the flow of a fluid. The curl tells us how much and in which direction the fluid is rotating around a point.

Mathematically, the curl is a vector that can be calculated using the cross product of the del operator and the vector field. For a vector field \( \mathbf{F} = P \mathbf{i} + Q \mathbf{j} + R \mathbf{k} \), the curl is given by:\[ abla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ P & Q & R \end{vmatrix} \]

By computing this determinant, you get a new vector field that describes the curl. In simple terms, if the curl is non-zero at a point, it means there is some rotation occurring at that point.

Line Integral

A line integral is a kind of definite integral. Instead of integrating over an interval on a number line, you integrate a function along a curve in space. This is especially useful in physics and engineering for finding work done by a force field.

In the context of vector fields, a line integral allows you to calculate the total effect of the vector field along a path. For a vector field \( \mathbf{F} \), and a parametrized curve \( \mathbf{r}(t) \), the line integral is:\[ \oint_C \mathbf{F} \cdot d\mathbf{r} = \int_a^b \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) \, dt \]

This integral sums up the component of the vector field that is tangential to the curve, which gives the accumulated effect along that curve in space.

Surface Integral

Surface integrals extend the concept of line integrals to surfaces. They are used to calculate quantities distributed over a surface, like the total flux of a field through a surface.

For a scalar function defined over a surface, the surface integral sums up the function's values at each point on the surface. For a vector field \( \mathbf{F} \) over a surface \( S \), the surface integral with the normal vector \( \mathbf{n} \) is:\[ \iint_S \mathbf{F} \cdot d\mathbf{S} \]

Here, \( d\mathbf{S} \) represents a tiny piece of the surface with a normal direction given by \( \mathbf{n} \). It's like adding up all the tiny contributions of the vector field coming out of the surface, which helps in applications like evaluating the flow of fluid or electric fields through a boundary.

Vector Field

A vector field is a function that assigns a vector to every point in space. These vectors describe quantities having both magnitude and direction, like wind speed and direction in weather forecasts.

Mathematically, a vector field \( \mathbf{F} \) is given as \( \mathbf{F}(x, y, z) = P(x, y, z) \mathbf{i} + Q(x, y, z) \mathbf{j} + R(x, y, z) \mathbf{k} \), where \( P, Q, \) and \( R \) are functions that determine the components of the vector field.

Understanding vector fields is crucial in physics and engineering as they model forces, fluid flows, and other dynamic systems in multidimensional spaces. They help us visually understand how values change through these spaces, making them an essential part of vector calculus.