Problem 340 - Calculus Volume 3

Question

[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.

Step-by-Step Solution

Answer

The value of the integral is 0.

Step 1: Identify the Surface and Curve

The surface \( S \) is bounded by the curve \( C \), which is the intersection of the plane \( 3x+2y+z=6 \) and the cylinder \( x^2+y^2=4 \). The surface \( S \) can be visualized as a circular cap on the plane above the cylinder. The curve \( C \) is oriented clockwise when viewed from above, meaning it has a positive orientation in terms of the surface normal following Stokes' theorem.

Step 2: Apply Stokes' 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 around its boundary \( \partial S \), the curve \( C \). Mathematically, this is: \[ \iint_{S}(\operatorname{curl}\mathbf{F} \cdot \mathbf{N})\, d S = \oint_{C} \mathbf{F} \cdot d\mathbf{r}. \]

Step 3: Parameterize the Curve

Since the intersection is with the cylinder \( x^2 + y^2 = 4 \), we can parameterize the curve \( C \) as: \[ \begin{align*}x &= 2\cos t, \y &= 2\sin t, \z &= 6 - 3(2\cos t) - 2(2\sin t) = 6 - 6\cos t - 4\sin t,\end{align*} \]where \( t \) ranges from \( 0 \) to \( 2\pi \).

Step 4: Calculate the Line Integral

The line integral is given by: \[ \oint_{C} \mathbf{F} \cdot d\mathbf{r} = \int_0^{2\pi} \Big((2\cos t)^2 (2\sin t) \, (-2\sin t) + (2\cos t)(2\sin t)^2 \, (2\cos t) + 6^3 \, 0 \Big) \, dt. \]Simplifying, this integral reduces to: \[ \int_0^{2\pi} (8\cos^2 t \sin^2 t) \, dt \] alone, which calculates using CAS to \( 0 \).

Key Concepts

Curl of a Vector Field

The curl of a vector field is a fundamental concept in vector calculus. It is used to describe the rotation of a vector field. Think of it as a measure of how much and in what direction the field is swirling around at a given point. Mathematically, for a vector field \ \( \mathbf{F} = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k} \ \), the curl is calculated as: \[ \operatorname{curl} \mathbf{F} = abla \times \mathbf{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right)\mathbf{i} + \left( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} \right)\mathbf{j} + \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right)\mathbf{k}. \]
Here, \ \( abla \times \mathbf{F} \ \) indicates the curl, and the partial derivatives measure changes in the components of the vector field in relation to each coordinate axis.

• It is a vector itself, showing how rotation happens about each axis.
• In physical applications, it can represent things like the rotational speed of a fluid at a point.

Surface Integral

A surface integral is a way to integrate over a surface in three-dimensional space. It helps us determine the flow of a field through a surface. Basically, it extends the idea of an integral from curves to surfaces.When you have a vector field, the surface integral of the curl over surface \ \( S \ \) can be expressed as: \[ iint_{S} \operatorname{curl} \mathbf{F} \cdot \mathbf{N} \, dS \] where \ \( \mathbf{N} \ \) is the unit normal vector to surface \ \( S \ \), and \ \( dS \ \) is the differential area element of the surface.

• This integral provides information on the net rotation effect that the field contributes to the surface area.
• It involves both the magnitude and direction of the field's components relative to the surface.

In the exercise, using Stokes' Theorem, which connects surface integrals and line integrals, can make calculations feasible and insightful for vector fields.

Line Integral

A line integral, in essence, measures the influence of a vector field along a curve. It evaluates how much of the field's force acts along the path that the curve takes. For a vector field \ \( \mathbf{F} = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k} \ \) and a curve \ \( C \ \), the line integral is expressed as: \[ \oint_{C} \mathbf{F} \cdot d\mathbf{r} = \oint_{C} (P \, dx + Q \, dy + R \, dz) \] where \ \( d\mathbf{r} \ \) is a tiny segment along the curve.

• The integral essentially sums up the component of the vector field along the direction of the curve.
• This is particularly useful in physics for work done by a force field along a path.

In practical computations, especially in the given exercise, finding the line integral involves parameterizing the curve (next concept) and then evaluating the integral over the parameter range.

Parametrization of Curves

Parametrization is a crucial technique used to describe curves in a more manageable way, especially when dealing with integrals. It involves expressing each of the coordinates of a point on the curve as a function of a single parameter.In the exercise, the curve \ \( C \ \) is the intersection of a plane and a cylinder. The parameterization uses trigonometric functions because the intersection's path is circular—ideal for sine and cosine functions: \[ \begin{align*}x &= 2\cos t, \ y &= 2\sin t, \ z &= 6 - 6\cos t - 4\sin t\end{align*} \] with \ \( t \ \) ranging from \ \( 0 \ \) to \ \( 2\pi \ \).

• This parameterization helps convert the problem into a single-variable calculus problem.
• It simplifies calculations by turning the curve path into continuous components that can be easily integrated.

With this technique, evaluating line integrals or even visualizing the curve becomes simpler and systematic.