Problem 19 - Calculus Early Transcendentals: Pearson New International Edition

Question

Calculate \(\iint_{2 S} \mathbf{F} \cdot \mathbf{n} d S\) for each of the following. Looked at the right way, all are quite easy and some are even trivial. (a) \(\mathbf{F}=(2 x+y z) \mathbf{i}+3 y \mathbf{j}+z^{2} \mathbf{k}\); \(S\) is the solid sphere \(x^{2}+y^{2}+z^{2} \leq 1\). (b) \(\mathbf{F}=\left(x^{2}+y^{2}+z^{2}\right)^{5 / 3}(x \mathbf{i}+y \mathbf{j}+z \mathbf{k}) ; S\) as in part (a). (c) \(\mathbf{F}=x^{2} \mathbf{i}+y^{2} \mathbf{j}+z^{2} \mathbf{k}\) \(S\) is the solid sphere \((x-2)^{2}+y^{2}+z^{2} \leq 1\). (d) \(\mathbf{F}=x^{2} \mathbf{i} ; S\) is the cube \(0 \leq x \leq 1,0 \leq y \leq 1,0 \leq z \leq 1\). (e) \(\mathbf{F}=(x+z) \mathbf{i}+(y+x) \mathbf{j}+(z+y) \mathbf{k} ; S\) is the tetrahedron cut from the first octant by the plane \(3 x+4 y+2 z=12\). (f) \(\mathbf{F}=x^{3} \mathbf{i}+y^{3} \mathbf{j}+z^{3} \mathbf{k} ; S\) as in part (a). (g) \(\mathbf{F}=(x \mathbf{i}+y \mathbf{j}) \ln \left(x^{2}+y^{2}\right)\); \(S\) is the solid cylinder \(x^{2}+y^{2} \leq 4,0 \leq z \leq 2\).

Step-by-Step Solution

Answer

(a) \(\frac{20}{3}\pi\), (b) 0, (c) 0, (d) 1, (e) 72, (f) 0, (g) 0.

Step 1: Understanding the Divergence Theorem

The Divergence Theorem states that \( \iint_{S} \mathbf{F} \cdot \mathbf{n} \, dS = \iiint_{V} (abla \cdot \mathbf{F}) \, dV \) where \( S \) is the boundary surface of volume \( V \), \( \mathbf{F} \) is a vector field, and \( \mathbf{n} \) is the outward unit normal. We'll use this theorem to solve each part.

Step 2: Solving Part (a)

Given \( \mathbf{F}=(2 x+y z) \mathbf{i}+3 y \mathbf{j}+z^{2} \mathbf{k} \), calculate \( abla \cdot \mathbf{F} = \frac{\partial}{\partial x}(2x + yz) + \frac{\partial}{\partial y}(3y) + \frac{\partial}{\partial z}(z^2) = 2 + 3 + 2z = 5 + 2z \). Now integrate over the volume of the sphere, \( \iiint_{V} (5 + 2z) \, dV \), where \( V \) is the sphere \( x^2 + y^2 + z^2 \leq 1 \). By symmetry and because \( z \) over the entire sphere should cancel out, the integral of \( z \) is zero, so evaluate \( \iiint_{V} 5 \, dV \). The volume of the sphere is \( \frac{4}{3}\pi (1)^3 = \frac{4}{3} \pi \), thus giving us \( 5 \times \frac{4}{3} \pi = \frac{20}{3} \pi \).

Step 3: Solving Part (b)

Given \( \mathbf{F} = (x^2+y^2+z^2)^{5/3} (x\mathbf{i} + y\mathbf{j} + z\mathbf{k}) \). Since this vector field is a radial field and everywhere perpendicular to the level surfaces of \( x^2+y^2+z^2 \), applying the divergence theorem, notice \( abla \cdot \mathbf{F} = 0 \) everywhere inside the sphere. By symmetry, the net flow out of the sphere is zero. Thus, \( \iint_S \mathbf{F} \cdot \mathbf{n} \, dS = 0 \).

Step 4: Solving Part (c)

Compute \( abla \cdot \mathbf{F} = \frac{\partial}{\partial x}(x^2) + \frac{\partial}{\partial y}(y^2) + \frac{\partial}{\partial z}(z^2) = 2x + 2y + 2z \). The sphere is now centered at \((2,0,0)\), so symmetry does not make any variables zero. However, compute \( \iiint_{V} (2x + 2y + 2z) \, dV \) over the new volume defined by \( (x-2)^2+y^2+z^2 \leq 1 \). The symmetry simplifies parts of this integral. However, notice the integral of linear components like \( x, y, z \) over symmetrical limits results in zero again, hence the net result is still zero.

Step 5: Solving Part (d)

Here, \( \mathbf{F} = x^2 \mathbf{i} \), hence \( abla \cdot \mathbf{F} = \frac{\partial}{\partial x}(x^2) = 2x \), giving the total divergence. Calculate \( \iiint_{0}^{1} \iiint_{0}^{1} \iiint_{0}^{1} 2x \, dx \, dy \, dz \). The integral becomes \( 2 \times \int_{0}^{1} x \, dx = 2 \times \frac{1}{2} = 1 \), leading to the total outward flux equating to "1".

Step 6: Solving Part (e)

For \( \mathbf{F}=(x+z) \mathbf{i}+(y+x) \mathbf{j}+(z+y) \mathbf{k} \), compute \( abla \cdot \mathbf{F} = 1 + 1 + 1 = 3 \). Integrate over the volume of a tetrahedron defined by \( 3x + 4y + 2z = 12 \) in the first octant. The volume \( V = \frac{1}{3} \times \, base \, \times \, height = \frac{1}{3} \times \frac{12 \times 12}{2} = 24 \). The integral becomes \( 3 \times 24 = 72 \), hence the flux is "72".

Step 7: Solving Part (f)

For \( \mathbf{F}=x^3 \mathbf{i}+y^3 \mathbf{j}+z^3 \mathbf{k} \), compute the divergence \( abla \cdot \mathbf{F} = 3x^2 + 3y^2 + 3z^2 \), integrate over the unit sphere \( x^2 + y^2 + z^2 \leq 1 \). Use symmetry or spherical coordinates, noticing contributions cancel out, resulting in zero given the properties of sphere symmetry around the origin. Hence, the flux is "0".

Step 8: Solving Part (g)

For \( \mathbf{F}=(x \mathbf{i}+y \mathbf{j}) \ln{(x^2+y^2)} \), it's clear that \( \mathbf{F} \) depends only on \( x \) and \( y \). Thus, it has no divergence in \( z \), mainly because of symmetry along the cylinder axis. Therefore, using the divergence theorem in cylindrical coordinates, and recognizing that the divergence \( abla \cdot \mathbf{F} \) is \(0\) within the entire volume of the cylinder \(x^2+y^2 \leq 4, 0 \leq z \leq 2\), the net outward flux is "0" given symmetry along the boundary in the \( xy \) plane.

Key Concepts

Flux Integrals

Flux integrals are central to understanding the flow of vector fields through surfaces. Imagine you have a surface, like a soap bubble, and you're interested in how much fluid passes through this bubble. The flux integral quantifies this.
Flux is calculated using the surface integral of a vector field over a given surface. Mathematically, it is represented as \( \iint_{S} \mathbf{F} \cdot \mathbf{n} \, dS \), where \( \mathbf{F} \) is the vector field, \( \mathbf{n} \) is the unit normal vector at a point on the surface, and \( S \) is the surface.

The challenge often lies in computing these integrals directly. But using handy theorems, such as the Divergence Theorem, simplifies this task by relating it to a volume integral.
Understanding flux integrals helps in fields like electromagnetism and fluid dynamics, where you need to know the rate at which something flows through a surface.

Vector Fields

Vector fields can be visualized as a collection of vectors spread out in a space, much like arrows pointing in different directions with various magnitudes. A vector field \( \mathbf{F} \) is defined such that each point in the space is assigned a vector. These fields offer incalculable insight into the behavior of physical systems, such as airflow over wings or electric fields around charged particles.

Mathematically, vector fields are expressed using components, typically along the x, y, and z axes, for three-dimensional spaces. For example, if \( \mathbf{F} = P \mathbf{i} + Q \mathbf{j} + R \mathbf{k} \), then at each point \((x, y, z)\), the vector \((P, Q, R)\) represents the vector field.

Approaching problems with vector fields often requires techniques like differentiation, specifically computing divergence, which determines how much a field behaves like a source or sink at a point.

Surface Integrals

Surface integrals are a way to generalize the idea of integration over flat surfaces to curved surfaces. Surface integrals enable the calculation of functions across surfaces that could be like spheres, cylinders, or irregular shapes. They are a cornerstone in physical applications needing a measure of flow or flux across a surface.

The general form of a surface integral of a vector field \( \mathbf{F} \) over a surface \( S \) is \( \iint_{S} \mathbf{F} \cdot \mathbf{n} \, dS \), where \( \mathbf{n} \) signifies the surface's unit normal vector.

Surface integrals are commonly used in the Divergence Theorem, particularly relating to integrating over closed surfaces that form the boundary of a volume. This linkage between surface and volume integrals enables sophisticated problem-solving strategies.

Volume Integrals

Volume integrals expand on the concept of integrating functions over regions in 3D space rather than lines or surfaces. It's akin to finding the 'sum' of a function's values over a 3D territory, facilitating calculations in diverse areas like mass distribution calculations or electromagnetics.

Typically, a volume integral takes the form \( \iiint_{V} \rho \, dV \), where \( \rho \) is a function over the volume \( V \). For instance, if we're interested in the divergence of a vector field, the volume integral usually considers this product.

Combining volume integrals with theorems like the Divergence Theorem transforms complex surface integral problems into more manageable volume-encapsulated calculations, providing a powerful toolset in mathematical physics and engineering.