Chapter 14: Problem 10
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
Step by step solution
Parameterize the Curve C
Compute \(\vec{F} \cdot d\vec{r} \) for Line Integral
Evaluate the Line Integral over C
Calculate \( \operatorname{curl} \vec{F} \)
Set Up the Surface Integral
Parameterize Surface S
Evaluate the Surface Integral
Conclusion of Stokes' Theorem Verification
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
vector fields
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.
line integrals
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 \( C \).
surface integrals
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.
curl of a vector 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.