Problem 4
Question
Verify Formula (2) in Stokes’ Theorem by evaluating the line integral and the surface integral. Assume that the surface has an upward orientation.. $$ \begin{array}{l}{\mathbf{F}(x, y, z)=(z-y) \mathbf{i}+(z+x) \mathbf{j}-(x+y) \mathbf{k} ; \sigma \text { is the por- }} \\ {\text { tion of the paraboloid } z=9-x^{2}-y^{2} \text { above the } x y \text { -plane. }}\end{array} $$
Step-by-Step Solution
Verified Answer
The line integral and surface integral both evaluate to \(18\pi\), verifying Stokes' Theorem.
1Step 1: Set up for Line Integral
First, determine the boundary curve for the surface \(\sigma\). Since \(\sigma\) is the portion of the paraboloid \(z=9-x^2-y^2\) above the \(xy\)-plane, the boundary curve occurs when \(z=0\), thus forming a circle \(x^2 + y^2 = 9\). Parametrize this boundary as \(\mathbf{r}(t) = (3 \cos t) \mathbf{i} + (3 \sin t) \mathbf{j}\), with \(t\) ranging from 0 to \(2\pi\).
2Step 2: Compute the Line Integral
Substitute the parametric equations into \(\mathbf{F}\): \(\mathbf{F}(x,y,z) = (-9\cos t - 3\sin t)\mathbf{i} + (-3\cos t + 9 \sin t)\mathbf{j} - 3 \cos t \mathbf{k}\). Compute the line integral \(\oint_C \mathbf{F} \cdot d\mathbf{r}\), where \(d\mathbf{r} = (-3\sin t) \mathbf{i} + (3\cos t) \mathbf{j}\) dt, as \(\oint (9\cos t + 3\sin t)^2 + (3\cos t - 9\sin t)^2 \ dt\), from 0 to \(2\pi\).
3Step 3: Simplify the Line Integral
Evaluate the integral by integrating each component separately and using trigonometric identities: \[\oint_C (9\cos t + 3\sin t)^2 dt + \oint_C (3\cos t - 9\sin t)^2 dt\]. Simplifying gives \[0\], because the result of each term integrating separately over \([0, 2\pi]\) results in 0.
4Step 4: Set up for Surface Integral
Now for the surface integral, take the curl of \(\mathbf{F}\), compute \(abla \times \mathbf{F}\). Calculate \(abla \times \mathbf{F} = (-2 \mathbf{i} + 2 \mathbf{j} + 2 \mathbf{k})\). The surface \(\sigma\) is parameterized by \(x\) and \(y\) with \(z=9-x^2-y^2\).
5Step 5: Compute Surface Integral
Choose an upward orientation for normal \(\mathbf{n}\) calculated using \(\mathbf{n} = (-2x\mathbf{i} - 2y\mathbf{j} + \mathbf{k})\). Compute \(\iint_\sigma (abla \times \mathbf{F}) \cdot \mathbf{n} \, dS\), which equals to \(\int_{x^2+y^2 \leq 9} 2(1) \,dA\).
6Step 6: Finish Surface Integral Calculation
Convert the region of integration to polar coordinates: \(x = r \cos \theta, y = r \sin \theta\) with \(0 \leq r \leq 3\) and \(0 \leq \theta \leq 2\pi\), giving \(\iint_{0}^{2\pi} \int_{0}^{3} 2\, r \,dr \,d\theta\). Evaluating gives \(18\pi\).
7Step 7: Conclusion: Verify Equality
Both integrals result in a value of \(18\pi\). Therefore, the surface integral over \(\sigma\) is equal to the line integral around its boundary, consistent with Stokes' Theorem.
Key Concepts
Line IntegralSurface IntegralParameterization
Line Integral
The line integral is an essential concept in vector calculus that allows for the integration of vector fields along a curve. When considering a line integral, we often focus on a vector field and a path or curve over which we integrate. In the context of Stokes' Theorem, the line integral operates over the boundary of a surface. This is key when confirming the equality implied by Stokes' Theorem.
To perform a line integral, we first need to parameterize the curve. For example, if we have a circular boundary like in our paraboloid example, we can use trigonometric functions for parameterization: \( \mathbf{r}(t) = (3 \cos t) \mathbf{i} + (3 \sin t) \mathbf{j} \, over \, t \, from \, 0 \, to \, 2\pi \,\). This effectively describes the circular path in terms of a single variable \(t\).
Once parameterized, the curve's direction and length contribute to evaluating the line integral. Calculating this involves substituting the parameterization into the vector field \(\mathbf{F}\), deriving \(d\mathbf{r}\) from \(\mathbf{r}(t)\), and performing the integral over the parameter's interval. This process transforms the line integral into the evaluation of an ordinary integral, simplified using trigonometric identities. In our solution, careful calculation using these methods reveals that the line integral evaluates to zero.
To perform a line integral, we first need to parameterize the curve. For example, if we have a circular boundary like in our paraboloid example, we can use trigonometric functions for parameterization: \( \mathbf{r}(t) = (3 \cos t) \mathbf{i} + (3 \sin t) \mathbf{j} \, over \, t \, from \, 0 \, to \, 2\pi \,\). This effectively describes the circular path in terms of a single variable \(t\).
Once parameterized, the curve's direction and length contribute to evaluating the line integral. Calculating this involves substituting the parameterization into the vector field \(\mathbf{F}\), deriving \(d\mathbf{r}\) from \(\mathbf{r}(t)\), and performing the integral over the parameter's interval. This process transforms the line integral into the evaluation of an ordinary integral, simplified using trigonometric identities. In our solution, careful calculation using these methods reveals that the line integral evaluates to zero.
Surface Integral
Surface integrals extend the concept of single-variable integration to functions over surfaces. Similar to using line integrals for curves, surface integrals are useful for understanding properties distributed over an entire surface. They play a pivotal role in Stokes' Theorem, which links a surface integral of a curl of a vector field over a surface to a line integral over its boundary.
To compute a surface integral, we need a parameterization of the surface. In our exercise, the surface is the top half of a paraboloid, described by \( z = 9 - x^2 - y^2 \). We first find the curl \( abla \times \mathbf{F} \) to understand how the vector field rotates over the surface. This curl gets integrated across the surface with respect to a chosen orientation.
Choosing an appropriate normal vector for the surface is crucial because it affects the orientation of the integral. Typically, an upward normal is used for surfaces above the xy-plane. With these pieces in place, the surface integral can be computed by switching to polar coordinates for easier integration. This involves substituting \( x = r \cos \theta \) and \( y = r \sin \theta \), leading to a more straightforward evaluation process. In our scenario, the calculated value of the surface integral confirmed Stokes' Theorem by equating it to the line integral.
To compute a surface integral, we need a parameterization of the surface. In our exercise, the surface is the top half of a paraboloid, described by \( z = 9 - x^2 - y^2 \). We first find the curl \( abla \times \mathbf{F} \) to understand how the vector field rotates over the surface. This curl gets integrated across the surface with respect to a chosen orientation.
Choosing an appropriate normal vector for the surface is crucial because it affects the orientation of the integral. Typically, an upward normal is used for surfaces above the xy-plane. With these pieces in place, the surface integral can be computed by switching to polar coordinates for easier integration. This involves substituting \( x = r \cos \theta \) and \( y = r \sin \theta \), leading to a more straightforward evaluation process. In our scenario, the calculated value of the surface integral confirmed Stokes' Theorem by equating it to the line integral.
Parameterization
Parameterization is a method used to represent a curve or surface in terms of one or more variables, effectively simplifying complex geometric and calculus problems. For a line, we often use a single variable, while for more complex surfaces or volumes, multiple parameters might be needed.
In the problem we're solving, parameterization is key both for the circle forming the boundary of the surface and the surface itself. For the circle, a simple parameterization using trigonometric functions \( \mathbf{r}(t) = (3 \cos t) \mathbf{i} + (3 \sin t) \mathbf{j} \) allows us to express the entire circular path in terms of the variable \( t \). This reduces the complexity of integration by allowing all calculations to be performed in one dimension.
For the paraboloid surface, parameterization enhances the calculation of the surface integral by switching from Cartesian to polar coordinates \((r, \theta)\). This streamlines the integration process and helps manage the surface's curved nature. Parameterization is indispensable in vector calculus as it converts multidimensional problems into more manageable forms, making the integration process more intuitive.
In the problem we're solving, parameterization is key both for the circle forming the boundary of the surface and the surface itself. For the circle, a simple parameterization using trigonometric functions \( \mathbf{r}(t) = (3 \cos t) \mathbf{i} + (3 \sin t) \mathbf{j} \) allows us to express the entire circular path in terms of the variable \( t \). This reduces the complexity of integration by allowing all calculations to be performed in one dimension.
For the paraboloid surface, parameterization enhances the calculation of the surface integral by switching from Cartesian to polar coordinates \((r, \theta)\). This streamlines the integration process and helps manage the surface's curved nature. Parameterization is indispensable in vector calculus as it converts multidimensional problems into more manageable forms, making the integration process more intuitive.
Other exercises in this chapter
Problem 3
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