The plane S can be parameterized by r(u, v) = (u, v, 3-u-2v with (u, v) E D, where D is bounded by u = 0, v = 0, and the line >
The plane S can be parameterized by r(u, v) = (u, v, 3-u-2v with (u, v) E D, where D is bounded by u = 0, v = 0, and the line >
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![The plane \( S \) can be parameterized by \(\vec{r}(u, v) = \langle u, v, 3 - u - 2v \rangle\)
with \( (u, v) \in D \), where \( D \) is bounded by \( u = 0 \), \( v = 0 \), and the line [incomplete expression].](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F09d105f3-6c69-4cbc-8997-6988f1733e6f%2Fc220ca9c-49b7-4428-b704-908d74af586a%2F1ilna4o_processed.png&w=3840&q=75)
Transcribed Image Text:The plane \( S \) can be parameterized by \(\vec{r}(u, v) = \langle u, v, 3 - u - 2v \rangle\)
with \( (u, v) \in D \), where \( D \) is bounded by \( u = 0 \), \( v = 0 \), and the line [incomplete expression].
![To evaluate the surface integral of the curl of a vector field \(\mathbf{F}\) over a surface \(S\), we relate it to a double integral over a domain \(D\):
\[
\iint_S \operatorname{curl}(\mathbf{F}) \cdot d\mathbf{S} = \iint_D \operatorname{curl}(\mathbf{F}(\mathbf{r}(u, v))) \cdot (\mathbf{r}_u \times \mathbf{r}_v) dA,
\]
where \( \mathbf{r}_u \times \mathbf{r}_v \) represents the cross product of partial derivatives of the parameterization \(\mathbf{r}(u, v)\).
To solve this, we need to determine \(\operatorname{curl}(\mathbf{F})\) and \(\mathbf{r}_u \times \mathbf{r}_v\):
1. The curl of \(\mathbf{F}\) is given by:
\[
\operatorname{curl}(\mathbf{F}) = \langle 1, 2, -1 \rangle
\]
2. The cross product \(\mathbf{r}_u \times \mathbf{r}_v\) is:
\[
\mathbf{r}_u \times \mathbf{r}_v = \langle 1, 2, 1 \rangle
\]
Finally, substitute these into the double integral:
\[
\iint_D \operatorname{curl}(\mathbf{F}(\mathbf{r}(u, v))) \cdot (\mathbf{r}_u \times \mathbf{r}_v) dA = \iint_D 4 \, dA
\]
This integral evaluates to 4 times the area of \(D\), where \(dA\) is the differential area element over the domain \(D\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F09d105f3-6c69-4cbc-8997-6988f1733e6f%2Fc220ca9c-49b7-4428-b704-908d74af586a%2Ff1meqdr_processed.png&w=3840&q=75)
Transcribed Image Text:To evaluate the surface integral of the curl of a vector field \(\mathbf{F}\) over a surface \(S\), we relate it to a double integral over a domain \(D\):
\[
\iint_S \operatorname{curl}(\mathbf{F}) \cdot d\mathbf{S} = \iint_D \operatorname{curl}(\mathbf{F}(\mathbf{r}(u, v))) \cdot (\mathbf{r}_u \times \mathbf{r}_v) dA,
\]
where \( \mathbf{r}_u \times \mathbf{r}_v \) represents the cross product of partial derivatives of the parameterization \(\mathbf{r}(u, v)\).
To solve this, we need to determine \(\operatorname{curl}(\mathbf{F})\) and \(\mathbf{r}_u \times \mathbf{r}_v\):
1. The curl of \(\mathbf{F}\) is given by:
\[
\operatorname{curl}(\mathbf{F}) = \langle 1, 2, -1 \rangle
\]
2. The cross product \(\mathbf{r}_u \times \mathbf{r}_v\) is:
\[
\mathbf{r}_u \times \mathbf{r}_v = \langle 1, 2, 1 \rangle
\]
Finally, substitute these into the double integral:
\[
\iint_D \operatorname{curl}(\mathbf{F}(\mathbf{r}(u, v))) \cdot (\mathbf{r}_u \times \mathbf{r}_v) dA = \iint_D 4 \, dA
\]
This integral evaluates to 4 times the area of \(D\), where \(dA\) is the differential area element over the domain \(D\).
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