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Chapter 18: Problem 14

Verify the divergence theorem (18.26) by evaluating both the surface integral and the triple integral. \(\mathbf{F}=\|\mathbf{r}\|^{2} \mathbf{r}\) for \(\mathbf{r}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\) \(S\) is the sphere \(x^{2}+y^{2}+z^{2}=a^{2}\)

Short Answer

Expert verified
Both integrals equal \(4 \pi a^4\), verifying the Divergence Theorem.

Step by step solution

01

Define the Vector Field and Sphere

The vector field is given as \( \mathbf{F} = \| \mathbf{r} \|^2 \mathbf{r} \), where \( \mathbf{r} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k} \). The sphere \( S \) is defined by the equation \( x^2 + y^2 + z^2 = a^2 \).
02

Surface Integral over the Sphere

For the surface integral, we use the formula \( \iint_S \mathbf{F} \cdot \mathbf{n} \, dS \), where \( \mathbf{n} \) is the outward normal vector. On a sphere of radius \( a \), \( \mathbf{n} \) is \( \frac{\mathbf{r}}{a} \), and \( dS = a^2 \sin\theta \, d\theta \, d\phi \). Substitute \( \mathbf{F} = a^2 \mathbf{r} \), and the integral becomes \[ a^4 \iint_S \sin\theta \, d\theta \, d\phi. \] This evaluates to \( 4\pi a^4 \), the surface area of the sphere multiplied by \( a^2 \).
03

Calculating the Divergence of the Vector Field

Calculate the divergence \( abla \cdot \mathbf{F} \). We know \( \mathbf{F} = \| \mathbf{r} \|^2 \mathbf{r} = (x^2 + y^2 + z^2)(x \mathbf{i} + y \mathbf{j} + z \mathbf{k}) \). By computing each partial derivative, we find \( abla \cdot \mathbf{F} = 5(x^2 + y^2 + z^2) \).
04

Triple Integral over the Volume of the Sphere

To find the volume integral, evaluate \(\iiint_V (abla \cdot \mathbf{F}) \, dV\). Substitute \( abla \cdot \mathbf{F} = 5r^2 \) and convert to spherical coordinates. The volume integral becomes \[ 5 \int_0^{2\pi} \int_0^\pi \int_0^a r^4 \sin\theta \, dr \, d\theta \, d\phi. \] After evaluating, this also results in \( 4 \pi a^4 \).
05

Compare the Results of the Integrals

Both the surface integral and the volume integral yield the result \( 4 \pi a^4 \). Thus, the Divergence Theorem, which states that these two integrals should be equal, is verified.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vector Field
A vector field is essentially a map that assigns a vector to every point in space. You can imagine it as a way of depicting directions and magnitudes throughout an area. In our exercise, the vector field is defined as \( \mathbf{F} = \| \mathbf{r} \|^2 \mathbf{r} \), where \( \mathbf{r} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k} \). This expression indicates that each vector in the field points in the direction of \( \mathbf{r} \) and is scaled by the square of the magnitude of \( \mathbf{r} \).

To break it down even further:
  • \( \mathbf{r} \) represents a position vector from the origin to a point \( (x, y, z) \).
  • \( \| \mathbf{r} \|^2 \) equals \( x^2 + y^2 + z^2 \), which is the square of the distance from the origin to that point.
  • The vector field \( \mathbf{F} \) thus changes based on location, direction, and its distance from the origin.
Understanding vector fields is pivotal when solving problems involving fields such as force, electric or magnetic fields, or flow velocity fields in fluids.
Surface Integral
Surface integrals are used to calculate the flux of a vector field through a surface. The flux represents how much of something passes through a surface in a vector field, like air passing through a net.

In the case of our problem, we are calculating the surface integral over the sphere defined by \( x^2 + y^2 + z^2 = a^2 \). The concept of the surface integral is employed using the formula:
  • \( \iint_S \mathbf{F} \cdot \mathbf{n} \, dS \)
  • Here, \( \mathbf{n} \) is the normal vector pointing outward from the surface, and for a sphere, \( \mathbf{n} = \frac{\mathbf{r}}{a} \).
  • The element of the area on the sphere, \( dS \), is given by \( a^2 \sin\theta \, d\theta \, d\phi \).
The result from this calculation is a value representing how much of the vector field permeates through the surface of the sphere. This value was found to be \( 4\pi a^4 \) in our example.
Triple Integral
Triple integrals extend the concept of integration to three dimensions. They are essential for evaluating the integral over a volume in space, often required for finding total quantities like mass or charge within solid objects.

In our example, we use a triple integral to compute the total divergence of the vector field within the volume of a sphere. The divergence \( abla \cdot \mathbf{F} \) is first determined, producing a function that measures how much a vector field spreads out from each point.

For our vector field, we calculate the divergence as \( 5(x^2 + y^2 + z^2) \).
To find the volume integral over this, spherical coordinates are an efficient choice:
  • The triple integral becomes \( 5 \int_0^{2\pi} \int_0^\pi \int_0^a r^4 \sin\theta \, dr \, d\theta \, d\phi \).
  • This integral ultimately evaluates to \( 4 \pi a^4 \), exactly the same as the result from the surface integral, thus confirming our use of the Divergence Theorem.
Spherical Coordinates
Spherical coordinates offer a seamless way to describe points in three-dimensional space, especially when dealing with spheres or radial symmetry. Essentially, they use three parameters:
  • \( r \): the radial distance from the origin to a point, akin to the radius of a sphere.
  • \( \theta \): the angle from the positive z-axis, similar to latitude.
  • \( \phi \): the angle from the positive x-axis in the xy-plane, akin to longitude.
These coordinates are particularly convenient for the type of symmetries present in the sphere \( x^2 + y^2 + z^2 = a^2 \) as seen in our example. The transformation from Cartesian to spherical coordinates is pivotal for simplifying integrals as it captures the symmetry of spheres naturally.

In the context of our solution:
  • The volume element \( dV \) in spherical coordinates translates to \( r^2 \sin\theta \, dr \, d\theta \, d\phi \); this accounts for the geometry of a sphere.
  • This transformation makes the triple integral tractable, especially when combined with the function forms derived from divergence expressions.
Mastering spherical coordinates can greatly simplify integrals involving spheres and similar bodies.

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Most popular questions from this chapter

Find the flux of \(F\) over the closed surface S. (Use the outer normal to S.) \(\mathbf{F}(x, y, z)=(x+y) \mathbf{i}+z \mathbf{j}+x z \mathbf{k}\) \(S\) is the surface of the cube having vertices \((\pm 1,\pm 1,\pm 1) .\)

Let \(\mathbf{F}\) be an inverse square field such that \(\mathbf{F}(x, y, z)=\) \(\left(c /\|\mathbf{r}\|^{3}\right) \mathbf{r}\) for \(\mathbf{r}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k},\) where \(c\) is a constant. Let \(P_{1}\) and \(P_{2}\) be any points whose distances from the origin are \(d_{1}\) and \(d_{2}\), respectively. Express, in terms of \(d_{1}\) and \(d_{2},\) the work done by \(\mathbf{F}\) along any piecewise-smooth curve joining \(P_{1}\) to \(P_{2}\).

Use Green's theorem to evaluate the line integral. \(\oint_{c}\left(1-x^{2} y\right) d x+\sin y d y\) \(C\) is the boundary of the region that lies inside the square with vertices (卤2,卤2) and outside the square with vertices (卤1,卤1) .

Assume that \(S\) and \(Q\) satisfy the conditions of the divergence theorem and that \(f\) and \(g\) are scalar functions that have continuous second partial derivatives. Prove the identity. \(\iiint_{Q}\left(f \nabla^{2} g+\nabla f \cdot \nabla g\right) d V=\iint_{S}(f \nabla g) \cdot \mathbf{n} d S\)

Use Green's theorem to evaluate the line integral. \(\oint_{c} x y d x+\sin y d y\) \(C\) is the triangle with vertices (1,1),(2,2),(3,0)
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