Chapter 18: Problem 18
Find the flux of \(F\) over the closed surface S. (Use the outer normal to S.) $$ \mathbf{F}(x, y, z)=x \mathbf{i}-y \mathbf{j}+z \mathbf{k} $$ \(S\) is the surface of the solid bounded by the graphs of \(z=x^{2}+y^{2}\) and \(z=4\)
Short Answer
Expert verified
The flux is \(8\pi\).
Step by step solution
01
Understanding the Geometry of Surface S
The surface, \(S\), is a solid bound between two surfaces: \(z = x^2 + y^2\) (a paraboloid) and \(z = 4\) (a plane). The region it encloses is a paraboloid capped at \(z=4\).
02
Applying Gauss's Divergence Theorem
According to Gauss's Divergence Theorem, the flux of \(\mathbf{F}\) across a closed surface \(S\) is equal to the triple integral of the divergence of \(\mathbf{F}\) over the volume \(V\) enclosed by \(S\). Mathematically, \(\iint_S \mathbf{F} \cdot \mathbf{n} \,dS = \iiint_V abla \cdot \mathbf{F} \,dV\).
03
Calculating the Divergence of F
First, calculate the divergence of \(\mathbf{F}\). We have \(abla \cdot \mathbf{F} = \frac{\partial}{\partial x}(x) + \frac{\partial}{\partial y}(-y) + \frac{\partial}{\partial z}(z) = 1 - 1 + 1 = 1\). Thus, \(abla \cdot \mathbf{F} = 1\).
04
Setting Up the Integral for Volume V
The volume, \(V\), is the region between \(z = x^2 + y^2\) and \(z = 4\). Convert to cylindrical coordinates for convenience, where \(x = r \cos \theta\), \(y = r \sin \theta\), and \(z = z\). The bounds are: \(0 \leq r \leq 2\), \(0 \leq \theta \leq 2\pi\), and \(r^2 \leq z \leq 4\).
05
Evaluating the Volume Integral
The integral is set as \(\iiint_V 1 \, dV = \int_{0}^{2\pi} \int_{0}^{2} \int_{r^2}^{4} 1 \, dz \, r \, dr \, d\theta\). Evaluate the innermost integral: \(\int_{r^2}^{4} 1 \, dz = 4 - r^2\). This leads to the middle integral \(\int_{0}^{2} (4 - r^2) r \, dr\).
06
Solving the Radial Integral
Simplify and calculate: \(\int_{0}^{2} (4 - r^2) r \, dr = \int_{0}^{2} (4r - r^3) \, dr = \left[ 2r^2 - \frac{r^4}{4} \right]_0^2 = 8 - 4 = 4\).
07
Final Integration over Theta
Finally, evaluate the integral over \(\theta\): \(\int_{0}^{2\pi} 4 \, d\theta = 4[\theta]_0^{2\pi} = 4(2\pi) = 8\pi\).
08
Conclusion: Finding the Total Flux
The total flux of \(\mathbf{F}\) across the surface \(S\) is \(8\pi\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Gauss's Divergence Theorem
When you need to calculate the flux of a vector field across a closed surface, Gauss's Divergence Theorem is a powerful tool.
This theorem helps us convert a surface integral into a volume integral, which can make calculations much simpler. The theorem states that the flux of a vector field \( \mathbf{F} \) through a closed surface \( S \) is equal to the triple integral of the divergence of \( \mathbf{F} \) over the volume \( V \) enclosed by \( S \). In mathematical terms, this is:
This theorem helps us convert a surface integral into a volume integral, which can make calculations much simpler. The theorem states that the flux of a vector field \( \mathbf{F} \) through a closed surface \( S \) is equal to the triple integral of the divergence of \( \mathbf{F} \) over the volume \( V \) enclosed by \( S \). In mathematical terms, this is:
- \( \iint_S \mathbf{F} \cdot \mathbf{n} \,dS = \iiint_V abla \cdot \mathbf{F} \,dV \)
- The left side is the surface integral, which involves the dot product of the vector field and the outward normal.
- The right side involves calculating the divergence, \( abla \cdot \mathbf{F} \), across the volume.
cylindrical coordinates
When dealing with symmetrical shapes, such as those with circular symmetry, cylindrical coordinates can be a very useful tool. Unlike the standard Cartesian coordinates \((x, y, z)\), cylindrical coordinates incorporate a radial distance \(r\), an angle \(\theta\), and a height \(z\).To convert from Cartesian to cylindrical coordinates, use the following formulas:
- \( x = r \cos \theta \)
- \( y = r \sin \theta \)
- \( z = z \)
- \(0 \leq r \leq 2\)
- \(0 \leq \theta \leq 2\pi\)
- \(r^{2} \leq z \leq 4\)
vector fields
Vector fields describe a vector value assigned to every point in space. A vector field like \( \mathbf{F}(x, y, z) = x \mathbf{i} - y \mathbf{j} + z \mathbf{k} \) gives us a sense of how vectors vary throughout space. They appear in physics and engineering, representing things like force fields, velocity fields, and more. In this problem, the vector field involves:
- An \( x \)-component: \( x \mathbf{i} \)
- A \( y \)-component: \( -y \mathbf{j} \)
- A \( z \)-component: \( z \mathbf{k} \)
- \( \frac{\partial}{\partial x}(x) = 1 \)
- \( \frac{\partial}{\partial y}(-y) = -1 \)
- \( \frac{\partial}{\partial z}(z) = 1 \)