Problem 14 - 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. $$ \begin{aligned} &\mathbf{F}=y \mathbf{i}+(y-x) \mathbf{j}+z^{2} \mathbf{k} ; S \text { that portion of the sphere }\\\ &x^{2}+y^{2}+(z-4)^{2}=25 \text { for } z \geq 0 \end{aligned} $$

Step-by-Step Solution

Answer

The integral evaluates to \(-50\pi\).

Step 1: Understand Stokes' Theorem

Stokes' Theorem relates a surface integral of a curl over a surface S to a line integral around the boundary curve C of S. Mathematically, it is stated as \( \iint_{S} (\text{curl} \mathbf{F}) \cdot \mathbf{n} \, dS = \oint_{C} \mathbf{F} \cdot d\mathbf{r} \). Here, \( \mathbf{n} \) is the unit normal to the surface S.

Step 2: Identify Surface S and Its Boundary

The given surface S is the top hemisphere (since \( z \geq 0 \)) of the sphere \( x^2 + y^2 + (z-4)^2 = 25 \). The boundary of S, which we will denote as C, is a circle in the plane \( z = 0 \) with center at \((0, 0, 4)\) and radius 5.

Step 3: Parametrize the Boundary Curve C

The boundary C is a circle of radius 5 in the xy-plane at \( z = 0 \). We can parametrize C as \( \mathbf{r}(t) = 5\cos(t) \mathbf{i} + 5\sin(t) \mathbf{j} + 0\mathbf{k} \), where \( t \) ranges from 0 to \( 2\pi \).

Step 4: Compute the Line Integral

Now, compute \( \oint_{C} \mathbf{F} \cdot d\mathbf{r} \). First, substitute \( \mathbf{r}(t) \) into \( \mathbf{F} \). We get \( \mathbf{F} = 5\sin(t) \mathbf{i} + (5\sin(t) - 5\cos(t)) \mathbf{j} + 0\mathbf{k} \). The differential \( d\mathbf{r} = -5\sin(t) \mathbf{i} + 5\cos(t) \mathbf{j} \, dt \).

Step 5: Evaluate the Integral

Compute the dot product \( \mathbf{F} \cdot d\mathbf{r} = (5\sin(t))(-5\sin(t)) + (5\sin(t) - 5\cos(t))(5\cos(t)) \). Simplify to obtain \( -25\sin^2(t) + 25\sin(t)\cos(t) - 25\cos^2(t) \). Further simplify using the identity \( \sin^2(t) + \cos^2(t) = 1 \) to get \( -25 \). Integrate over \( t \) from 0 to \( 2\pi \): \( \int_0^{2\pi} (-25) \, dt = -25[2\pi - 0] = -50\pi \).

Step 6: Conclusion

According to Stokes' Theorem, the value of \( \iint_{S} (\text{curl} \mathbf{F}) \cdot \mathbf{n} \, dS \) is equal to the line integral, which we calculated as \( -50\pi \).

Key Concepts

Surface Integrals

Surface integrals are a way to sum up a function over a surface, much like a regular integral sums over an interval. The difference is that, in a surface integral, the function may be a scalar field or a vector field. In the context of vector fields, which is what Stokes' Theorem uses, the surface integral often involves the dot product of a vector field and the normal vector to the surface.

When applying surface integrals in Stokes' Theorem, you deal with the curl of a vector field across a surface. To compute this, you set up the integral of the curl of the vector field, denoted as \( \iint_{S} (\text{curl} \mathbf{F}) \cdot \mathbf{n} \, dS \), where \( \mathbf{n} \) is the unit normal to the surface. For example, if the vector field relates to a flow of air or liquid, the surface integral can represent the total flow across the surface.

In our exercise, \( S \) is the upper hemisphere of a sphere, and its 'upward' orientation means that the normal vector \( \mathbf{n} \) points outward, away from the center. The integration gives the overall 'rotation' of the vector field across the surface.

Line Integrals

Line integrals involve integrating a function along a curve. In the context of vector fields, they often take the form \( \oint_{C} \mathbf{F} \cdot d\mathbf{r} \), where \( \mathbf{F} \) is the vector field and \( d\mathbf{r} \) is an infinitesimal vector along the curve \( C \). Line integrals can be thought of as measuring the 'work' done by the field along the path of the curve.

In relation to Stokes' Theorem, the line integral is around the boundary of the surface over which the surface integral is taken. So, for a surface \( S \) with a boundary curve \( C \), the theorem connects the two integrals, stating that the surface integral of the curl over \( S \) equals the line integral around \( C \).

In our exercise, the boundary curve \( C \) is a circle lying at \( z = 0 \) in the xy-plane, exemplifying a simple loop for calculation. The parametric representation \( \mathbf{r}(t) = 5\cos(t) \mathbf{i} + 5\sin(t) \mathbf{j} \) assists in evaluating the line integral by substituting it back into the vector field \( \mathbf{F} \). By doing this, one can easily follow along and compute \( \oint_{C} \mathbf{F} \cdot d\mathbf{r} \).

The step-by-step evaluation of this integral is crucial in showing that the surface integral and line integral truly match, as Stokes' Theorem states.

Vector Fields

A vector field assigns a vector to every point in space, often used to represent physical quantities like velocity, gravitational forces, or electromagnetic fields. In mathematics and physics, understanding vector fields is crucial for analyzing and interpreting directions and magnitudes across regions.

In our task, the vector field \( \mathbf{F} = y \mathbf{i} + (y-x) \mathbf{j} + z^2 \mathbf{k} \) is provided over a spherical surface. The vector field describes how vectors 'flow' or 'rotate' around space. For instance, the \( y \mathbf{i} \) component suggests movement or an effect aligned with the x-axis that depends on the y-coordinate, while \( z^2 \mathbf{k} \) indicates a flow aligned through the z-axis that strengthens quadratically as z increases.

To apply Stokes' Theorem, one must first calculate the curl of the vector field, which measures the tendency to rotate around a point. By using Stokes' theorem, we convert evaluating the potentially complex surface behavior by examining the simpler problem of a line integral around the boundary. This can be extremely useful in physics and engineering contexts, simplifying the process of calculating circulation or rotational effects.