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Chapter 5: Problem 14

[T] Use a CAS and Stokes' theorem to evaluate \(\iint_{S}(\operatorname{curl} \mathbf{F} \cdot \mathbf{N}) d S\), where \(\mathbf{F}(x, y, z)=x^{2} y \mathbf{i}+x y^{2} \mathbf{j}+z^{3} \mathbf{k}\) and \(C\) is the curve of the intersection of plane \(3 x+2 y+z=6\) and cylinder \(x^{2}+y^{2}=4\), oriented clockwise when viewed from above.

Short Answer

Expert verified
The value of the surface integral is zero using Stokes' theorem.

Step by step solution

01

Understanding Stokes鈥� Theorem

Stokes' Theorem relates the surface integral of the curl of a vector field over a surface \(S\) to the line integral of the vector field over the boundary curve \(C\) of \(S\). Mathematically, it is expressed as: \(\iint_{S}(\operatorname{curl} \mathbf{F} \cdot \mathbf{N}) d S = \oint_{C} \mathbf{F} \cdot \mathbf{dr}\). Our task is to find \(\oint_{C} \mathbf{F} \cdot \mathbf{dr}\).
02

Parametrize the Curve C

Since \(C\) is the intersection of the plane \(3x + 2y + z = 6\) and the cylinder \(x^2 + y^2 = 4\), we first parametrize the cylinder. In the xy-plane, this circle can be represented as \((x, y) = (2 \cos t, 2 \sin t)\) for \(0 \leq t < 2\pi\). Substitute into the plane equation to get \(z = 6 - 3x - 2y = 6 - 6\cos t - 4\sin t\). Thus, the curve can be parametrized by \(r(t) = (2\cos t, 2\sin t, 6 - 6\cos t - 4\sin t)\).
03

Calculate dr

Find the derivative of \(r(t)\) with respect to \(t\) for use in the line integral. \(r'(t) = \left( -2\sin t, 2\cos t, 6\sin t - 4\cos t \right)\).
04

Evaluate F along C

Along the curve \(C\), substitute the parameterization into \(\mathbf{F}\). This gives: \(\mathbf{F}(r(t)) = \left((2 \cos t)^2 (2 \sin t), (2 \cos t) (2 \sin t)^2, (6 - 6 \cos t - 4 \sin t)^3\right)\). Simplify each component: \(= (8\cos^2t\sin t, 8\cos t\sin^2t, (6 - 6\cos t - 4\sin t)^3)\).
05

Setup Line Integral

The line integral \(\oint_{C} \mathbf{F} \cdot \mathbf{dr}\) becomes \(\int_{0}^{2\pi} \left[8\cos^2t\sin t (-2\sin t) + 8\cos t\sin^2t (2\cos t) + (6 - 6\cos t - 4\sin t)^3 (6\sin t - 4\cos t)\right] dt\).
06

Compute the Integral

Use a computer algebra system (CAS) to compute \(\int_{0}^{2\pi} \left[-16\cos^2t\sin^2t + 16\cos^2t\sin^2t + (6 - 6\cos t - 4\sin t)^3 (6\sin t - 4\cos t)\right] dt\). Simplify the expression and calculate the integral, which results in zero after symmetry considerations and simplification.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Curl of a Vector Field
The curl of a vector field is a measure of the rotation of the field around a point. For a vector field \( \mathbf{F} = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k} \), the curl is given by:
  • \( abla \times \mathbf{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right) \mathbf{i} + \left( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} \right) \mathbf{j} + \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \mathbf{k} \)
This essentially provides a vector that summarizes how the field curls or "twists" around each point. In terms of Stokes' Theorem, the curl plays a vital role in converting the problem from a surface integral to a more manageable line integral along a curve.
Parametrization of Curves
Parametrization is a way to represent a curve using a parameter, usually \( t \), which helps simplify complex expressions and computations. For example, given a curve defined by the intersection of a plane and a cylinder,
  • We use parameters like \((x, y, z) = (2\cos t, 2\sin t, 6 - 6\cos t - 4\sin t)\)
This representation makes it easier to handle calculations, such as determining the derivative with respect to \( t \), which provides insights into direction and rapid computations needed for integrals.
Line Integrals
Line integrals are used to compute a scalar field or vector field along a curve. When dealing with a vector field \( \mathbf{F} \) and a parametrized curve \( \mathbf{r}(t) \), the line integral is expressed as:
  • \( \oint_{C} \mathbf{F} \cdot \mathbf{dr} = \int_{a}^{b} \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) \, dt \)
This integral calculates the "work done" by the vector field along the curve \( C \). For areas where the path depends on a complex geometry, the integration simplifies to evaluating the vectors along the path parametrized over \( t \). This powerful tool, complemented by Stokes' Theorem, converts a problem in 3D surfaces into something solvable with simpler line computations.
Surface Integrals
Surface integrals extend the idea of line integrals over a surface. They evaluate how a vector field like \( \operatorname{curl} \mathbf{F} \) behaves over a surface \( S \).
  • The surface integral \( \iint_{S} (\operatorname{curl} \mathbf{F} \cdot \mathbf{N}) \, dS \) computes the total "twist" of the vector field over \( S \).
This operation is crucial in physical applications, such as fluid dynamics or electromagnetism, where understanding the aggregate behavior over a surface helps in grasping the system's full dynamics. Stokes' Theorem elegantly bridges this with line integrals, providing a dual perspective useful in vector calculus and applied mathematics.

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