Chapter 6: Problem 342
ITI Use a CAS and Stokes' theorem to evaluate \(\iint_{S} \operatorname{curl} \mathbf{F} \cdot d \mathbf{S}\) where \(\mathbf{F}(x, y, z)=z^{2} \mathbf{i}-3 x y \mathbf{j}+x^{3} y^{3} \mathbf{k}\) and \(S\) is the top part of \(z=5-x^{2}-y^{2}\) above plane \(z=1\) and \(S\) is oriented upward.
Short Answer
Expert verified
The integral evaluates to 0.
Step by step solution
01
Understanding the Context
We are asked to evaluate \(\iint_{S} \operatorname{curl} \mathbf{F} \cdot d \mathbf{S}\), where \(\mathbf{F}(x, y, z) = z^{2} \mathbf{i} - 3xy \mathbf{j} + x^{3}y^{3} \mathbf{k}\) over the surface \(S\), which is the top part of the paraboloid \(z = 5 - x^2 - y^2\) above plane \(z=1\) with upward orientation. We will apply Stokes' theorem to transform this surface integral to a line integral.
02
Stokes' Theorem
Stokes' Theorem relates the surface integral of the curl of a vector field over a surface \(S\) to a line integral over the boundary \(\partial S\) of that surface. It states that \(\iint_{S} \operatorname{curl} \mathbf{F} \cdot d \mathbf{S} = \oint_{\partial S} \mathbf{F} \cdot d\mathbf{r}\). Our task is to find this line integral around the boundary of \(S\).
03
Define the Boundary
The boundary \(\partial S\) is the circle \(z = 1\) on the plane where the surface meets it. Thus, it can be described in polar coordinates as \(x^{2} + y^{2} = 4\), a circle of radius 2 around the origin.
04
Parametrize the Boundary
To evaluate the line integral, we parametrize the circle \(x = 2\cos\theta, y = 2\sin\theta, z = 1\), where \(\theta\) ranges from 0 to \(2\pi\). The differential arc length is \(d\mathbf{r} = (\frac{dx}{d\theta}, \frac{dy}{d\theta}, \frac{dz}{d\theta})d\theta = (-2\sin\theta, 2\cos\theta, 0)d\theta\).
05
Evaluate the Line Integral
Substitute the parameterization into \(\mathbf{F}(x, y, z)\) and compute the dot product. We get \(\mathbf{F}(2\cos\theta, 2\sin\theta, 1) = \mathbf{i} - 12\cos\theta\sin\theta \mathbf{j} + 8\cos^{3}\theta\sin^{3}\theta \mathbf{k}\). Calculate the line integral: \[ \oint_{\partial S} \mathbf{F} \cdot d\mathbf{r} = \int_0^{2\pi} [(1)(-2\sin\theta) - (12\cos\theta\sin\theta)(2\cos\theta)]d\theta \]Simplify and integrate with respect to \(\theta\). After calculation, the result is 0.
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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 concept of the curl of a vector field helps us understand the rotation at a point in the field. It's often used in physics and engineering, especially in fluid dynamics and electromagnetism.
To compute the curl, we apply the abla (nabla) operator in a cross-product with the vector field \( \mathbf{F} \). In three-dimensional space, if \( \mathbf{F}(x, y, z) = P \mathbf{i} + Q \mathbf{j} + R \mathbf{k} \), the curl is given by:\[\operatorname{curl} \mathbf{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right) \mathbf{i} + \left( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} \right) \mathbf{j} + \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \mathbf{k}\]
To compute the curl, we apply the abla (nabla) operator in a cross-product with the vector field \( \mathbf{F} \). In three-dimensional space, if \( \mathbf{F}(x, y, z) = P \mathbf{i} + Q \mathbf{j} + R \mathbf{k} \), the curl is given by:\[\operatorname{curl} \mathbf{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right) \mathbf{i} + \left( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} \right) \mathbf{j} + \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \mathbf{k}\]
- Rotation: The curl represents the tendency to rotate around a point.
- Zero Curl: If the curl is zero, the field is irrotational.
Surface Integral
Surface integrals extend the idea of a double integral to integrating over a surface. It measures properties like flux through a surface.
For a vector field \( \mathbf{F} \), the surface integral over a surface \( S \) is written as:\[\iint_S \mathbf{F} \cdot d\mathbf{S}\]- \( d\mathbf{S} \) is the infinitesimal oriented surface area vector, based on the orientation of the surface.- For a surface defined parametrically, \( d\mathbf{S} \) can be determined using the cross-product of derivatives of the parametrization.
Surface integrals are crucial in problems involving flux, like electric and magnetic fields through a membrane, or in our case, helping bridge into Stokes' Theorem.
For a vector field \( \mathbf{F} \), the surface integral over a surface \( S \) is written as:\[\iint_S \mathbf{F} \cdot d\mathbf{S}\]- \( d\mathbf{S} \) is the infinitesimal oriented surface area vector, based on the orientation of the surface.- For a surface defined parametrically, \( d\mathbf{S} \) can be determined using the cross-product of derivatives of the parametrization.
Surface integrals are crucial in problems involving flux, like electric and magnetic fields through a membrane, or in our case, helping bridge into Stokes' Theorem.
Line Integral
A line integral allows us to integrate over a curve, finding quantities like work done by a force field along a path.
For a vector field \( \mathbf{F} \), it's expressed as:\[\oint_{C} \mathbf{F} \cdot d\mathbf{r}\]where \( C \) is a curve, and \( d\mathbf{r} \) is the differential vector along the curve.
For a vector field \( \mathbf{F} \), it's expressed as:\[\oint_{C} \mathbf{F} \cdot d\mathbf{r}\]where \( C \) is a curve, and \( d\mathbf{r} \) is the differential vector along the curve.
- Application: Great for work/energy problems in physics.
- Path Dependency: Generally depends on the path, unless the field is conservative.
Parametrization
Parametrization is the process of expressing a curve or surface in terms of parameters, making complex shapes easier to handle.
Imagine translating a circle from its standard form \( x^2 + y^2 = r^2 \) to a parameter form, using:\[x = r \cos\theta, \quad y = r \sin\theta\]where \( \theta \) ranges from 0 to \( 2\pi \).
- Allows structures that are difficult to work with algebraically to be handled smoothly.- Essential for fields like computer graphics and physics simulations, where defining shapes and surfaces explicitly isn't feasible.
In the context of Stokes' Theorem, parametrization helps convert a surface into a form where both surface and line integrals can be evaluated efficiently.
Imagine translating a circle from its standard form \( x^2 + y^2 = r^2 \) to a parameter form, using:\[x = r \cos\theta, \quad y = r \sin\theta\]where \( \theta \) ranges from 0 to \( 2\pi \).
- Allows structures that are difficult to work with algebraically to be handled smoothly.- Essential for fields like computer graphics and physics simulations, where defining shapes and surfaces explicitly isn't feasible.
In the context of Stokes' Theorem, parametrization helps convert a surface into a form where both surface and line integrals can be evaluated efficiently.