Problem 13 - 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. \(\mathbf{F}=6 y z \mathbf{i}+5 x \mathbf{j}+y z e^{x^{2}} \mathbf{k} ; S\) that portion of the paraboloid \(z=\frac{1}{4} x^{2}+y^{2}\) for \(0 \leq z \leq 4\)

Step-by-Step Solution

Answer

The value is \(-152\pi\).

Step 1: Understand the Problem Setup

We need to use Stokes' Theorem to evaluate \(\iint_{S}(\operatorname{curl} \mathbf{F})\cdot \mathbf{n} \, dS\). The surface \(S\) is part of the paraboloid \(z = \frac{1}{4}x^2 + y^2\) capped with \(0 \leq z \leq 4\). Since Stokes' Theorem relates surface integrals of curl to line integrals, we need to determine the boundary curve of the surface and the vector field.

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

Calculate the curl of \(\mathbf{F} = 6yz\mathbf{i} + 5x\mathbf{j} + yz e^{x^2}\mathbf{k}\). Using \((b, c, f)\) for the components of \(\mathbf{F}\), we have:\[ \operatorname{curl} \mathbf{F} = \left( \frac{\partial f}{\partial y} - \frac{\partial c}{\partial z} \right)\mathbf{i} + \left( \frac{\partial a}{\partial z} - \frac{\partial f}{\partial x} \right)\mathbf{j} + \left( \frac{\partial c}{\partial x} - \frac{\partial b}{\partial y} \right)\mathbf{k} \] The curl is: \( \operatorname{curl} \mathbf{F} = x ze^{x^2} \mathbf{i} + (z - 2xye^{x^2}) \mathbf{j} + 5\mathbf{k} \).

Step 3: Identify the Boundary Curve

The boundary curve, \(C\), of the surface \(S\) is the circle at the top of the paraboloid when \(z=4\). Plug \(z = 4\) into \(z = \frac{1}{4}x^2 + y^2\), simplifying gives \(x^2 + 4y^2 = 16\). This ellipse represents the boundary curve \(C\).

Step 4: Set Up Line Integral Using Stokes' Theorem

Stokes' Theorem states:\[ \iint_{S}(\operatorname{curl} \mathbf{F})\cdot \mathbf{n} \, dS = \oint_{C} \mathbf{F} \cdot d\mathbf{r} \]Parametrize the ellipse \(C\) using \(x = 4 \cos \theta, \, y = 2 \sin \theta\). The parameterization is closed and \(d\mathbf{r} = (-4 \sin \theta) \, d\theta\mathbf{i} + (2 \cos \theta) \, d\theta\mathbf{j}\).

Step 5: Calculate the Line Integral

Substitute \(x = 4\cos\theta\), \(y = 2\sin\theta\), and \(z = 4\) into \(\mathbf{F}\):\[ \mathbf{F}|_{C} = 48 \sin \theta\mathbf{i} + 20 \cos \theta\mathbf{j} + 32e^{16}\sin\theta\mathbf{k} \]Then compute:\[ \oint_{C} \mathbf{F} \cdot d\mathbf{r} = \int_{0}^{2\pi} (48 \sin \theta (-4 \sin \theta) + 20 \cos \theta (2 \cos \theta)) \, d\theta \]Simplify and evaluate the integral:\[ = \int_{0}^{2\pi} ( -192 \sin^2 \theta + 40 \cos^2 \theta) \, d\theta \]

Step 6: Simplify and Integrate

Use trigonometric identities \( \sin^2 \theta = \frac{1 - \cos 2\theta}{2}, \quad \cos^2 \theta = \frac{1 + \cos 2\theta}{2} \) to simplify: \[ \int_{0}^{2\pi} (-192 \frac{1 - \cos 2\theta}{2} + 40 \frac{1 + \cos 2\theta}{2}) \, d\theta \] This becomes: \[ \int_{0}^{2\pi} (-96 + 96\cos 2\theta + 20 - 20\cos 2\theta) \, d\theta = \int_{0}^{2\pi} (-76 + 76 \cos 2\theta) \, d\theta \] Integrating from 0 to \(2\pi\), considering the periodic terms cancel, results in \(-76 \times 2\pi\) = \(-152\pi\).

Key Concepts

Vector Calculus

Vector calculus is a branch of mathematics focused on analyzing vector fields. Vector fields are functions that assign a vector to every point in space, and they're often used to model physical phenomena like velocity, force, and magnetic fields. In vector calculus, we work with various operators and concepts to describe the behavior and properties of these fields.

Key operations include:

• Gradient: This operation takes a scalar field and produces a vector field, indicating the direction of the greatest rate of increase of the scalar.
• Divergence: This operation takes a vector field and returns a scalar field, measuring how much the field spreads out from a point.
• Curl: This operator, which we'll look at more closely later, maps a vector field to another vector field, reflecting the rotation at each point.

Understanding these operations is essential for solving problems in physics and engineering, where vector calculus is widely applied.

Surface Integrals

Surface integrals extend the concept of double integrals to functions defined over surfaces in three-dimensional space. In essence, they allow you to sum up quantities over a curved surface, rather than flat regions like in standard integrations.

When performing a surface integral of a vector field, one typically evaluates the integral:\[ \iint_{S} \mathbf{F} \cdot \mathbf{n} \, dS \]where:

• \( \mathbf{F} \) is the vector field.
• \( \mathbf{n} \) is the unit normal vector for the surface \( S \).
• \( dS \) is the differential element of the surface area.

This allows for calculation of quantities like the flow across a surface, where considering the directionality via \( \mathbf{n} \) is crucial. Surface integrals are foundational for applying Stokes’ Theorem, connecting the behavior over a surface to its boundary.

Line Integrals

Line integrals are a way of integrating functions along a curve. This is particularly useful in physics for calculating work done by a force along a path, among other applications.

The line integral of a vector field \( \mathbf{F} \) along a curve \( C \) is formulated as:\[ \oint_{C} \mathbf{F} \cdot d\mathbf{r} \]where:

• \( d\mathbf{r} \) is a differential vector element along the curve.

Stokes' theorem particularly shines in converting a surface integral into a line integral along the boundary curve of the surface. This conversion simplifies complex calculations, turning a problem with potentially intricate surfaces into manageable integrations along a curve.

Curl of a Vector Field

The curl of a vector field is a measure of the field's tendency to rotate around a point. Think of it as quantifying how much 'spin' or rotational movement the field exhibits.

Mathematically, the curl of a vector field \( \mathbf{F} = (F_1, F_2, F_3) \) in three dimensions is given by:\[ \operatorname{curl} \mathbf{F} = abla \times \mathbf{F} = \left( \frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z} \right) \mathbf{i} + \left( \frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x} \right) \mathbf{j} + \left( \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \right) \mathbf{k} \]

The curl can be visualized using a rotating paddle in a fluid: wherever the paddle spins, that's where curl is present. Learning to compute and interpret curl helps in recognizing how fields circulate, which is pivotal for applying the fundamental theorem underlying vector calculus, such as Stokes' Theorem.