Use the divergence theorem to find the inward flux of the vector field F = 3x²y²1+ yj - 6xy² zk through the boundary of the solid bounded by the paraboloid z = x² + y² and the plane z = 2y. Hint: Use cylindrical coordinates, and note that cos20 = (1 + cos20)/2, sin²0 = (1-cos20)/2
Use the divergence theorem to find the inward flux of the vector field F = 3x²y²1+ yj - 6xy² zk through the boundary of the solid bounded by the paraboloid z = x² + y² and the plane z = 2y. Hint: Use cylindrical coordinates, and note that cos20 = (1 + cos20)/2, sin²0 = (1-cos20)/2
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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![### Problem Q9
**Objective:**
Use the divergence theorem to find the inward flux of the vector field
\[ \mathbf{F} = 3x^2y^2\mathbf{i} + y\mathbf{j} - 6xy^2z\mathbf{k} \]
through the boundary of the solid bounded by the paraboloid
\[ z = x^2 + y^2 \]
and the plane
\[ z = 2y. \]
**Hint:**
Use cylindrical coordinates, and note that
\[ \cos^2\theta = \frac{1 + \cos 2\theta}{2}, \]
\[ \sin^2 \theta = \frac{1 - \cos 2\theta}{2}. \]
**Solution Approach:**
1. **Divergence Theorem Setup:**
- The divergence theorem relates the flux of a vector field through a closed surface to the volume integral of the divergence of the vector field over the region enclosed by the surface.
\[ \iint_{\partial V} \mathbf{F} \cdot \mathbf{n} \, dS = \iiint_V (\nabla \cdot \mathbf{F}) \, dV. \]
2. **Compute the Divergence \( \nabla \cdot \mathbf{F} \):**
- Calculate the partial derivatives:
\[ \nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(3x^2y^2) + \frac{\partial}{\partial y}(y) + \frac{\partial}{\partial z}(-6xy^2z). \]
3. **Region Description in Cylindrical Coordinates:**
- Convert the given surfaces to cylindrical coordinates where:
\[ x = r\cos \theta, \, y = r\sin \theta, \, z = z. \]
The paraboloid \( z = x^2 + y^2 \) becomes \( z = r^2 \), and the plane \( z = 2y \) becomes \( z = 2r\sin \theta \).
4. **Volume Integral Setup:**
- Boundaries for the integral in cylindrical coordinates:
- Radial coordinate \( r \) ranges from 0 to the value where \( z \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd3485286-e420-456a-8a41-5bc62e18ff5f%2Febe0ef6b-6758-41d3-9c0a-0212ecdc0ab3%2F3c9bb4y_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Q9
**Objective:**
Use the divergence theorem to find the inward flux of the vector field
\[ \mathbf{F} = 3x^2y^2\mathbf{i} + y\mathbf{j} - 6xy^2z\mathbf{k} \]
through the boundary of the solid bounded by the paraboloid
\[ z = x^2 + y^2 \]
and the plane
\[ z = 2y. \]
**Hint:**
Use cylindrical coordinates, and note that
\[ \cos^2\theta = \frac{1 + \cos 2\theta}{2}, \]
\[ \sin^2 \theta = \frac{1 - \cos 2\theta}{2}. \]
**Solution Approach:**
1. **Divergence Theorem Setup:**
- The divergence theorem relates the flux of a vector field through a closed surface to the volume integral of the divergence of the vector field over the region enclosed by the surface.
\[ \iint_{\partial V} \mathbf{F} \cdot \mathbf{n} \, dS = \iiint_V (\nabla \cdot \mathbf{F}) \, dV. \]
2. **Compute the Divergence \( \nabla \cdot \mathbf{F} \):**
- Calculate the partial derivatives:
\[ \nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(3x^2y^2) + \frac{\partial}{\partial y}(y) + \frac{\partial}{\partial z}(-6xy^2z). \]
3. **Region Description in Cylindrical Coordinates:**
- Convert the given surfaces to cylindrical coordinates where:
\[ x = r\cos \theta, \, y = r\sin \theta, \, z = z. \]
The paraboloid \( z = x^2 + y^2 \) becomes \( z = r^2 \), and the plane \( z = 2y \) becomes \( z = 2r\sin \theta \).
4. **Volume Integral Setup:**
- Boundaries for the integral in cylindrical coordinates:
- Radial coordinate \( r \) ranges from 0 to the value where \( z \
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