The surface integral is rewritten incorrectly. F((x, y)) is incorrect. The tangent vectors T, and T, are incorrect. The normal vector N is incorrect. No errors exist in the work shown. Compute f F d.S for the given oriented surface. F = (y, z, x), plane 5x - 6y + z = 1, 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, upward-pointing normal (Express numbers in exact form. Use symbolic notation and fractions where needed.)
The surface integral is rewritten incorrectly. F((x, y)) is incorrect. The tangent vectors T, and T, are incorrect. The normal vector N is incorrect. No errors exist in the work shown. Compute f F d.S for the given oriented surface. F = (y, z, x), plane 5x - 6y + z = 1, 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, upward-pointing normal (Express numbers in exact form. Use symbolic notation and fractions where needed.)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![**Instructions:**
- Select the statement that describes the error in the provided work:
- ☐ The surface integral is rewritten incorrectly.
- ☐ \( \mathbf{F}(\Phi(x, y)) \) is incorrect.
- ☐ The tangent vectors \( \mathbf{T}_x \) and \( \mathbf{T}_y \) are incorrect.
- ☐ The normal vector \( \mathbf{N} \) is incorrect.
- ☐ No errors exist in the work shown.
**Problem:**
Compute \(\iint_{S} \mathbf{F} \, dS\) for the given oriented surface.
\[ \mathbf{F} = \langle y, z, x \rangle \]
Surface defined by the plane \(5x - 6y + z = 1\),
with boundaries \(0 \leq x \leq 1\), \(0 \leq y \leq 1\), and an upward-pointing normal vector.
*(Express numbers in exact form. Use symbolic notation and fractions where needed.)*
**Solution:**
\[
\iint_{S} \mathbf{F} \, dS = \_\_\_\_
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F77548912-c51c-4c9d-8b51-f3905a3bec75%2F183f6aec-d0be-41f1-af70-b7561189f180%2Fioc7y9_processed.png&w=3840&q=75)
Transcribed Image Text:**Instructions:**
- Select the statement that describes the error in the provided work:
- ☐ The surface integral is rewritten incorrectly.
- ☐ \( \mathbf{F}(\Phi(x, y)) \) is incorrect.
- ☐ The tangent vectors \( \mathbf{T}_x \) and \( \mathbf{T}_y \) are incorrect.
- ☐ The normal vector \( \mathbf{N} \) is incorrect.
- ☐ No errors exist in the work shown.
**Problem:**
Compute \(\iint_{S} \mathbf{F} \, dS\) for the given oriented surface.
\[ \mathbf{F} = \langle y, z, x \rangle \]
Surface defined by the plane \(5x - 6y + z = 1\),
with boundaries \(0 \leq x \leq 1\), \(0 \leq y \leq 1\), and an upward-pointing normal vector.
*(Express numbers in exact form. Use symbolic notation and fractions where needed.)*
**Solution:**
\[
\iint_{S} \mathbf{F} \, dS = \_\_\_\_
\]
![**Surface Integral Calculation for an Oriented Surface**
**Objective:** Compute the surface integral \(\iint_S \mathbf{F} \, dS\) for the given oriented surface.
**Given Surface:**
\[
\mathbf{F} = \langle y, z, x \rangle, \text{plane } 5x - 6y + z = 1, 0 \leq x \leq 1, 0 \leq y \leq 1, \text{upward-pointing normal}
\]
**Procedure:**
1. **Partial Derivatives:**
To find \(\mathbf{T}_x\), compute the partial derivative of \(\Phi\) with respect to \(x\):
\[
\mathbf{T}_x = \frac{\partial \Phi}{\partial x} = \frac{\partial}{\partial x}(x, y, 1 - 5x + 6y) = \langle 1, 0, -5 \rangle
\]
To find \(\mathbf{T}_y\), compute the partial derivative of \(\Phi\) with respect to \(y\):
\[
\mathbf{T}_y = \frac{\partial \Phi}{\partial y} = \frac{\partial}{\partial y}(x, y, 1 - 5x + 6y) = \langle 0, 1, 6 \rangle
\]
2. **Cross Product:**
Compute the cross product \(\mathbf{T}_x \times \mathbf{T}_y\) as follows:
\[
\mathbf{T}_x \times \mathbf{T}_y = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 & 0 & -5 \\
0 & 1 & 6
\end{vmatrix} = 5\mathbf{i} - 6\mathbf{j} + \mathbf{k} = \langle 5, -6, 1 \rangle
\]
3. **Normal Vector:**
Since the plane is oriented with an upward-pointing normal, the normal vector is:
\[
\mathbf{N} = \langle](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F77548912-c51c-4c9d-8b51-f3905a3bec75%2F183f6aec-d0be-41f1-af70-b7561189f180%2Fzkedgbet_processed.png&w=3840&q=75)
Transcribed Image Text:**Surface Integral Calculation for an Oriented Surface**
**Objective:** Compute the surface integral \(\iint_S \mathbf{F} \, dS\) for the given oriented surface.
**Given Surface:**
\[
\mathbf{F} = \langle y, z, x \rangle, \text{plane } 5x - 6y + z = 1, 0 \leq x \leq 1, 0 \leq y \leq 1, \text{upward-pointing normal}
\]
**Procedure:**
1. **Partial Derivatives:**
To find \(\mathbf{T}_x\), compute the partial derivative of \(\Phi\) with respect to \(x\):
\[
\mathbf{T}_x = \frac{\partial \Phi}{\partial x} = \frac{\partial}{\partial x}(x, y, 1 - 5x + 6y) = \langle 1, 0, -5 \rangle
\]
To find \(\mathbf{T}_y\), compute the partial derivative of \(\Phi\) with respect to \(y\):
\[
\mathbf{T}_y = \frac{\partial \Phi}{\partial y} = \frac{\partial}{\partial y}(x, y, 1 - 5x + 6y) = \langle 0, 1, 6 \rangle
\]
2. **Cross Product:**
Compute the cross product \(\mathbf{T}_x \times \mathbf{T}_y\) as follows:
\[
\mathbf{T}_x \times \mathbf{T}_y = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 & 0 & -5 \\
0 & 1 & 6
\end{vmatrix} = 5\mathbf{i} - 6\mathbf{j} + \mathbf{k} = \langle 5, -6, 1 \rangle
\]
3. **Normal Vector:**
Since the plane is oriented with an upward-pointing normal, the normal vector is:
\[
\mathbf{N} = \langle
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