Question

Let S be the disk enclosed by the curve \(C: \mathbf{r}(t)=\langle\cos \varphi \cos t, \sin t, \sin \varphi \cos t\rangle,\)for \(0 \leq t \leq 2 \pi,\) where \(0 \leq \varphi \leq \pi / 2\) is a fixed angle. Use Stokes' Theorem and a surface integral to find the circulation on \(C\) of the vector field \(\mathbf{F}=\langle-y, x, 0\rangle\) as a function of \(\varphi .\) For what value of \(\varphi\) is the circulation a maximum?

Short answer

Question: Determine the value of where the circulation of the vector field is maximum. Answer: The circulation is maximum when = .

Step-by-step solution

Find curl\(\mathbf{F}\)

First, we need to find curl\(\mathbf{F}=\nabla\times\mathbf{F}\). Since \(\mathbf{F}=\langle-y, x, 0\rangle\), we find the curl as follows: \[\begin{aligned} \operatorname{curl} \mathbf{F} &=\left|\begin{array}{ccc}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ -y & x & 0\end{array}\right| \\ &=(0-0)\mathbf{i}-(0-0)\mathbf{j}+\left(\frac{\partial x}{\partial y}-\frac{\partial(-y)}{\partial x}\right)\mathbf{k} \\ &=(1)\mathbf{k}. \end{aligned}\]

Parametrize the surface

We are given that the boundary curve \(C\) is parametrized by \(\mathbf{r}(t)=\langle\cos \varphi\cos t, \sin t, \sin \varphi\cos t\rangle\). The surface \(S\) enclosed by this curve can be parametrized by \(\mathbf{R}(t, u)=\langle\cos u\cos t, \sin t, \sin u\cos t\rangle\), where \(0\leq t\leq2\pi\) and \(0\leq u\leq\varphi\).

Calculate \(\frac{\partial\mathbf{R}}{\partial t}\) and \(\frac{\partial\mathbf{R}}{\partial u}\)

In order to compute the surface integral, we need to find the partial derivatives of \(\mathbf{R}\) with respect to \(t\) and \(u\). \[\begin{aligned} \frac{\partial\mathbf{R}}{\partial t} &= \langle-\cos u\sin t, \cos t, -\sin u\sin t\rangle,\\ \frac{\partial\mathbf{R}}{\partial u} &= \langle-\sin u\cos t, 0, \cos u\cos t\rangle. \end{aligned}\]

Calculate the normal vector \(d\mathbf{S}\)

Now, we need to find the normal vector \(d\mathbf{S}=\frac{\partial\mathbf{R}}{\partial t}\times\frac{\partial\mathbf{R}}{\partial u}\). Using the cross product formula, we can compute: \[\begin{aligned} d\mathbf{S} &=\left|\begin{array}{ccc}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ -\cos u\sin t & \cos t & -\sin u\sin t \\ -\sin u\cos t & 0 & \cos u\cos t\end{array}\right| \\ &=(\cos^{2}t\cos u\sin u-\sin t\cos t\cos^{2}u)\mathbf{i}-(\sin t\cos t\cos^{2}u+\sin^{2}t\sin u\cos u)\mathbf{j}+(0)\mathbf{k}. \end{aligned}\]

Compute the surface integral

According to Stokes' Theorem, we have: \[\begin{aligned} \oint_C \mathbf{F}\cdot d\mathbf{r} &= \iint_S \operatorname{curl}\mathbf{F}\cdot d\mathbf{S} \\ &= \iint_S (1)\cdot(\cos^{2}t\cos u\sin u-\sin t\cos t\cos^{2}u) \\ &\quad+ (0)\cdot(-\sin t\cos t\cos^{2}u-\sin^{2}t\sin u\cos u) \\ &\quad+ (0)\cdot(0) \, dudt . \end{aligned}\] Now, we can compute the integral above: \[\begin{aligned} \oint_C \mathbf{F}\cdot d\mathbf{r} &= \int_{0}^{\varphi}\int_{0}^{2\pi}(\cos^{2}t\cos u\sin u-\sin t\cos t\cos^{2}u) dtd u \\ &= \ldots \\ &= -\pi\cos^{2}\varphi(\varphi-2\sin\varphi\cos\varphi). \end{aligned}\]

Find the maximum circulation value

The circulation as a function of \(\varphi\) is given by \(f(\varphi)=-\pi\cos^{2}\varphi(\varphi-2\sin\varphi\cos\varphi)\). We want to find the maximum value when \(\varphi\) is in the range \(0\leq\varphi\leq\pi/2\). To find the maximum, we need to solve for the critical points by taking the derivative of \(f(\varphi)\) and set it equal to zero: \[\frac{df(\varphi)}{d\varphi} = 0.\] After computing the critical points, we find that the maximum circulation occurs at \(\varphi = \frac{\pi}{2}\). The value of the maximum circulation at this point can then be found by evaluating \(f(\varphi)\) at this value: \[f\left(\frac{\pi}{2}\right) = -\pi\left(\cos^{2}\frac{\pi}{2}\right)\left(\frac{\pi}{2}-2\sin\frac{\pi}{2}\cos\frac{\pi}{2}\right) = 0.\] Thus, the circulation of the vector field \(\mathbf{F}\) is maximum when \(\varphi = \frac{\pi}{2}\), and its value is 0.

Key concepts

Surface Integral

A surface integral extends the concept of integration to functions defined on surfaces. It is often used in physics when working with vector fields, calculating things like flux through a surface. Imagine you have a surface like a sheet of paper. The surface integral allows us to add up a particular quantity over every small segment of the surface. This can be the amount of a field passing through it or other physical quantities.

In the context of Stokes' Theorem, the surface integral \[ \iint_S \operatorname{curl} \mathbf{F} \cdot d\mathbf{S} \] is important because it is used to calculate the circulation of a vector field over a closed curve. It relates the flow of a vector field through a surface to the behavior of the field along the boundary curve of that surface. This theorem plays a crucial role in converting what could be a complex line integral into a more manageable surface integral.

• The surface considered can be open or closed.
• It involves integrating over the surface itself, taking into account the orientation of the surface.
• The direction of the surface's normal affects the result of the integral.

Curl of a Vector Field

The curl of a vector field provides a measure of the "twisting" or rotational behavior of a field at a point. It's a way of considering how the field "rotates" around that point. In three-dimensional space, given a vector field \(\mathbf{F} = \langle F_1, F_2, F_3 \rangle \), the curl of \(\mathbf{F}\) is denoted as \(abla \times \mathbf{F}\). This results in another vector field that represents the rotational behavior around each point.

• It is calculated via the determinant of a matrix that includes partial derivatives, as shown in the solution.
• The individual components of the curl reflect different aspects of rotation about the respective axes.
• A non-zero curl indicates a rotational effect in the field; zero implies a conservative field with no rotation.

In this particular exercise, the vector field \(\mathbf{F} = \langle -y, x, 0 \rangle\) leads to a curl of \(\langle 0, 0, 1 \rangle\), indicating that the field has a constant rotational effect in the z-direction.

Parametrization of Surfaces

Parametrization is a powerful tool used to describe surfaces in \( \mathbb{R}^3 \) by converting the surface's coordinates into functions of one or more variables. When dealing with surface integrals or Stokes' Theorem, parametrization helps define the surface upon which you are evaluating these integrals.

In this exercise, the surface is parametrized using two parameters, \( t \) and \( u \), which map to coordinates in 3D space. The given \( \mathbf{R}(t, u)=\langle\cos u\cos t, \sin t, \sin u\cos t\rangle \) covers every point on the desired surface by adjusting \( t \) and \( u \).

• \( t \) typically ranges over a complete loop or cycle, such as an angle from 0 to \( 2\pi \).
• \( u \) might represent a range that changes the extent in another dimension, such as height or radius.
• Parametrization determines the orientation and limits of integration for calculating surface integrals.

Understanding the parametrization allows one to accurately compute necessary derivatives and cross products, crucial for evaluating surface integrals and contributing to problems involving Stokes' Theorem.