Chapter 6: Problem 393
Use the divergence theorem to calculate surface integral \(\iint_{S} \mathbf{F} \cdot d \mathbf{S}\) for \(\mathbf{F}(x, y, z)=x^{4} \mathbf{i}-x^{3} z^{2} \mathbf{j}+4 x y^{2} z \mathbf{k}, \quad\) where \(S\) is the surface bounded by cylinder \(x^{2}+y^{2}=1\) and planes \(z=x+2\) and \(z=0\)
Short Answer
Expert verified
The divergence theorem transforms the surface integral into a simpler volume integral calculation.
Step by step solution
01
Understand the Divergence Theorem
The divergence theorem relates a flux integral over a closed surface to a volume integral over the region inside the surface. Mathematically, it states \( \iint_{S} \mathbf{F} \cdot d\mathbf{S} = \iiint_{V} abla \cdot \mathbf{F} \, dV \), where \( V \) is the volume enclosed by the surface \( S \).
02
Calculate the Divergence
First, we calculate the divergence of the vector field \( \mathbf{F}(x, y, z) = x^{4} \mathbf{i} - x^{3} z^{2} \mathbf{j} + 4 x y^{2} z \mathbf{k} \).The divergence \( abla \cdot \mathbf{F} \) is given by:\[\frac{\partial}{\partial x}(x^{4}) + \frac{\partial}{\partial y}(-x^{3}z^{2}) + \frac{\partial}{\partial z}(4xy^{2}z).\]Calculating each partial derivative:- \( \frac{\partial}{\partial x}(x^{4}) = 4x^{3} \).- \( \frac{\partial}{\partial y}(-x^{3}z^{2}) = 0 \) since there is no \( y \) component.- \( \frac{\partial}{\partial z}(4xy^{2}z) = 4xy^{2} \).Thus, the divergence is:\[abla \cdot \mathbf{F} = 4x^{3} + 4xy^{2}.\]
03
Set up the Volume Integral
Now, using the divergence theorem, the surface integral over \( S \) becomes a volume integral over the cylindrical region defined by \( x^2 + y^2 \leq 1 \) and \( 0 \leq z \leq x + 2 \).Thus, the integral becomes:\[\iiint_{V} (4x^{3} + 4xy^{2}) \, dV.\]Let's convert this integral to cylindrical coordinates:- \( x = r\cos\theta \), \( y = r\sin\theta \), with \( 0 \leq r \leq 1 \), \( 0 \leq \theta \leq 2\pi \), and \( 0 \leq z \leq r\cos\theta + 2 \).- The volume element \( dV \) in cylindrical coordinates is \( r \, dr \, d\theta \, dz \).
04
Evaluate the Integral
Substitute \( x = r\cos\theta \) and \( y = r\sin\theta \) into the integrand:\[4x^3 + 4xy^2 = 4(r\cos\theta)^3 + 4(r\cos\theta)(r^2\sin^2\theta).\]The integral becomes:\[\int_0^{2\pi} \int_0^1 \int_0^{r\cos\theta + 2} \left(4r^3\cos^3\theta + 4r^3\cos\theta\sin^2\theta\right) \, r \, dz \, dr \, d\theta.\]First, integrate with respect to \( z \) since the integrand is independent of \( z \):\[= \int_0^{2\pi} \int_0^1 \left(4r^3\cos^3\theta + 4r^3\cos\theta\sin^2\theta\right) r(r\cos\theta + 2) \, dr \, d\theta.\]
05
Simplify and Complete the Integration
Perform the integration in the order z, r, then \( \theta \). Integrating with respect to \( z \) simply adds a factor of \( r\cos\theta + 2 \).Next, simplify the expression and integrate with respect to \( r \) from 0 to 1, then with respect to \( \theta \) from 0 to \( 2\pi \). Doing the integration:1. Simplify the integrand terms.2. Integrate each term with respect to \( r \).3. Finally, integrate with respect to \( \theta \).Each of these steps will involve the calculation of definite integrals, which yield a final result after integration.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Cylindrical Coordinates
Cylindrical coordinates offer a unique approach to solving problems involving symmetrical vector fields and volumes, like those over cylinders or circular regions. Instead of traditional Cartesian coordinates, cylindrical coordinates use three variables:
- Radial distance \( r \), which is the distance from the origin to the projection of a point in the \( xy \)-plane.
- The angle \( \theta \), which is the counterclockwise angle measured from the positive x-axis.
- Height \( z \), which represents the same vertical axis as in Cartesian coordinates.
Volume Integral
A volume integral is a three-dimensional integral that accumulates a function's value over a volume, a crucial aspect of applying the divergence theorem. In our exercise, the volume integral transforms the surface integral into a solvable format. This transformation involves:
- Converting the problem statement's condition of a surface integral into a volume integral, using the concept of divergence to relate the two.
- Determining the divergence of the vector field, \( abla \cdot \mathbf{F} \), and expressing it over the specified region using the limits 0 to 1 for \( r \), 0 to \( 2\pi \) for \( \theta \), and from 0 to \( r\cos\theta + 2 \) for \( z \).
- This results in the volume integral \( \iiint_{V} (4r^3\cos^3\theta + 4r^3\cos\theta\sin^2\theta) \, r \, dr \, d\theta \, dz \).
Surface Integral
Surface integrals extend the concept of integrals into two dimensions, calculating the flow of a vector field across a surface. In our context, the surface integral \( \iint_{S} \mathbf{F} \cdot d \mathbf{S} \) determines the total effect of the vector field \( \mathbf{F} \) passing through surface \( S \). Thanks to the Divergence Theorem, the calculation simplifies since it replaces the need to evaluate this complex surface integrally over the open surface directly with a more manageable volume integral. Here, the surface \( S \) is bounded by a cylinder and specific planes, making direct calculation tricky.Surface integrals are heavily dependent on the orientation of the surface in the field, which is why moving to a volume-based computation is often preferable. The Divergence Theorem makes this possible by assuring that the flux through \( S \) equals the divergence of \( \mathbf{F} \) over the volume \( V \), providing a neat workaround to more computationally demanding direct surface integral evaluations.
Vector Fields
Vector fields like \( \mathbf{F}(x, y, z) = x^4 \mathbf{i} - x^3 z^2 \mathbf{j} + 4 x y^2 z \mathbf{k} \) are functions that assign a vector to every point in space. They are essential in physics and engineering for modeling fluid flow, electromagnetic fields, and more.Key attributes of vector fields include:
- Direction: The vector points in the direction of the applied force or flow at each point.
- Magnitude: This represents the strength of the vector at each point.