Question

In Exercises \(9-12,\) a closed curve \(C\) that is the boundary of a surface \(S\) is given along with a vector field \(\vec{F}\). Verify Stokes' Theorem on \(C ;\) that is, show \(\oint_{c} \vec{F} \cdot d \vec{r}=\iint_{S}(\operatorname{curl} \vec{F}) \cdot \vec{n} d S\). $$ \begin{aligned} &C \text { is the curve parametrized by } \vec{r}(t)=\left\langle\cos t, \sin t, e^{-1}\right\rangle\\\ &\text { and } \mathcal{S} \text { is the portion of } z=e^{-x^{2}-y^{2}} \text { enclosed by } c ; \vec{F}=\\\ &\langle-y, x, 1\rangle . \end{aligned} $$

Short answer

Stokes' Theorem is verified: both integrals are equal to \(2\pi\).

Step-by-step solution

Parameterize the Curve C

The curve \( C \) is given as \( \vec{r}(t) = \langle \cos t, \sin t, e^{-1} \rangle \) for \( t \) ranging from \( 0 \) to \( 2\pi \). This represents a circle in the \( xy \)-plane at height \( z = e^{-1} \).

Compute \(\vec{F} \cdot d\vec{r} \) for Line Integral

First, compute the derivative of \( \vec{r}(t) \) with respect to \( t \), which gives \( \vec{r}'(t) = \langle -\sin t, \cos t, 0 \rangle \). Next, substitute \( \vec{r}(t) \) into \( \vec{F} \) to get \( \vec{F} = \langle -\sin t, \cos t, 1 \rangle \). The expression for \( \vec{F} \cdot d\vec{r} \) becomes \( (-\sin t, \cos t, 1) \cdot (-\sin t, \cos t, 0) \). Expanding, we find \( \sin^2 t + \cos^2 t = 1 \).

Evaluate the Line Integral over C

Integrate \( 1 \) with respect to \( t \) from \( 0 \) to \( 2\pi \): \( \int_0^{2\pi} 1 \, dt = 2\pi \). Thus, the line integral \( \oint_C \vec{F} \cdot d\vec{r} = 2\pi \).

Calculate \( \operatorname{curl} \vec{F} \)

The curl of \( \vec{F} = \langle -y, x, 1 \rangle \) is calculated using \( abla \times \vec{F} \). Compute: \[ abla \times \vec{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ -y & x & 1 \end{vmatrix} = \langle 0, 0, 1 + 1 \rangle = \langle 0, 0, 2 \rangle. \]

Set Up the Surface Integral

Evaluate the surface integral \( \iint_S (abla \times \vec{F}) \cdot \vec{n} \, dS \) where \( \vec{n} = \langle 0, 0, 1 \rangle \) is the unit normal, upward since \( z = e^{-x^2-y^2} \) is oriented counter-clockwise. Thus, \( (abla \times \vec{F}) \cdot \vec{n} = 2 \).

Parameterize Surface S

The region \( S \) where \( z = e^{-x^2-y^2} \) is bounded by the circle \( x^2 + y^2 \leq 1 \). Hence, parameterize using polar coordinates: \( x = r \cos \theta, \ y = r \sin \theta \) with \( r \leq 1 \).

Evaluate the Surface Integral

Perform the surface integral: \( \iint_S 2 \, dS = \int_0^{2\pi} \int_0^1 2 \cdot r \, dr \, d\theta \), since \( dS = r \, dr \, d\theta \). This results in \[ 2 \int_0^{2\pi} \int_0^1 r \, dr \, d\theta = 2 \int_0^{2\pi} \left[ \frac{r^2}{2} \right]_0^1 \, d\theta = 2 \int_0^{2\pi} \frac{1}{2} \, d\theta = 2 \pi. \]

Conclusion of Stokes' Theorem Verification

Both integrals, the line integral and the surface integral, equal \( 2\pi \). Thus, Stokes' Theorem is verified: \[ \oint_{C} \vec{F} \cdot d\vec{r} = \iint_{S} (abla \times \vec{F}) \cdot \vec{n} \, dS = 2\pi. \]

Key concepts

vector fields

Vector fields can be imagined as a collection of arrows associated with certain points in a space. Each arrow has a direction and a magnitude, representing a vector. In more technical terms, a vector field assigns a vector to every point in a space, which can be two-dimensional or three-dimensional, depending on the context.

For instance, think about the wind pattern over a geographical region. The wind at any point can be represented by a vector showing the wind's direction and speed. Similarly, in our exercise, the vector field \( \vec{F} = \langle -y, x, 1 \rangle \) assigns a vector to every point in the three-dimensional space. The components of this vector \( -y \), \( x \), and \( 1 \) determine its direction and length.

• \( -y \) indicates how much the vector points in the negative direction along the y-axis.
• \( x \) shows its component along the x-axis.
• 1 shows a constant upward component along the z-axis.

Understanding vector fields is crucial as they form the basis for more complex operations like line integrals and surface integrals.

line integrals

Line integrals involve integrating a function along a curve or a path. They are used to measure the total effect of a vector field along a given path, similar to summing up forces along a trajectory.

In this exercise, we take the line integral around a closed path, which is the boundary of a surface. This path, \( C \), is parameterized by \( \vec{r}(t) = \langle \cos t, \sin t, e^{-1} \rangle \). It traces a circle in the \( xy \)-plane at a constant height \( z = e^{-1} \).

• We calculate \( \vec{F} \cdot d\vec{r} \), the dot product of the vector field with the derivative of \( \vec{r}(t) \).
• Integrating \( \vec{F} \cdot d\vec{r} \) from \( 0 \) to \( 2\pi \) gives the total effect of the vector field along path \( C \).

Line integrals are essential in physics for work calculations, where force is applied along a path.

surface integrals

Surface integrals extend the concept of line integrals to calculus over two-dimensional surfaces. They allow us to average quantities over curved surfaces in space.

For our scenario, Stokes' Theorem involves calculating a surface integral over \( S \), which follows the surface defined by \( z = e^{-x^2-y^2} \). This surface is enclosed by the path \( C \) and is parameterized using polar coordinates \( x = r \cos \theta \), \( y = r \sin \theta \) with \( r \leq 1 \).

• The surface is covered using polar coordinates due to its circular boundary.
• The surface integral \( \iint_S (abla \times \vec{F}) \cdot \vec{n} \, dS \) involves integrating the curl of the vector field over this surface.

Surface integrals are key in calculating flux across surfaces, aiding in various applications in electromagnetism and fluid dynamics.

curl of a vector field

The curl of a vector field is a measure of the rotation or swirling of the field at a point. It tells us how much, and in which direction, a small object would rotate if placed in the field.

To compute the curl, we use the cross-product of the del operator with the vector field \( abla \times \vec{F} \). In our problem, this is calculated as \( \langle 0, 0, 2 \rangle \), which means the curl is acting entirely in the z-direction.

• This vector \( \langle 0, 0, 2 \rangle \) suggests uniform rotation in the plane parallel to the xy-plane.
• The magnitude (2 in this case) indicates the strength of the rotation.

Curl plays a fundamental role in understanding fluid flow and electromagnetic fields, making it a pillar in vector calculus.