Question

Evaluate each of the integrals as either a volume integral or a surface integral, whichever is easier. \(\iint \mathbf{V} \cdot \mathbf{n} d \sigma\) if \(\mathbf{V}=x \cos ^{2} y \mathbf{i}+x z \mathbf{j}+z \sin ^{2} y \mathbf{k}\) over the surface of a sphere with center at the origin and radius 3.

Short answer

The value of the surface integral is \( 36 \pi \).

Step-by-step solution

Understand the Integral Type

This integral is a surface integral of the vector field \(\mathbf{V}=x \cos ^{2} y \mathbf{i}+x z \mathbf{j}+z \sin ^{2} y \mathbf{k}\) over the surface of a sphere. The goal is to determine which method (volume integral or surface integral) is easier to evaluate.

Apply the Divergence Theorem

The Divergence Theorem relates the flux of a vector field through a surface to a volume integral over the region inside the surface: \[\iint_{S} \mathbf{V} \cdot \mathbf{n} \, d\sigma = \iiint_{V} abla \cdot \mathbf{V} \, dV\]

Calculate the Divergence of \(\mathbf{V}\)

Compute the divergence \(abla \cdot \mathbf{V}\):\[abla \cdot \mathbf{V} = \frac{\partial}{\partial x}(x \cos^{2} y) + \frac{\partial}{\partial y}(x z) + \frac{\partial}{\partial z}(z \sin^{2} y)\ = \cos^{2} y + 0 + \sin^{2} y = 1.\]

Set Up the Volume Integral

Applying the divergence theorem, we need to evaluate the volume integral over the sphere of radius 3. Since the divergence is 1, the integral simplifies to the volume of the sphere:\[\iiint_{V} 1 \, dV.\]

Calculate the Volume of the Sphere

The volume of a sphere with radius \( r \) is given by the formula \(\frac{4}{3} \pi r^{3}\). For a radius of 3, the volume is:\[\frac{4}{3} \pi (3)^{3} = 36 \pi.\]

Conclude the Solution

Since the volume integral is equivalent to the surface integral by the Divergence Theorem, the value of the surface integral is \( 36 \pi \).

Key concepts

The Divergence Theorem

The Divergence Theorem is a powerful tool in vector calculus. It relates the flow of a vector field through a closed surface to the behavior of the vector field inside the surface. Simply put, it allows you to convert a surface integral into a volume integral, which can be easier to evaluate.
Mathematically, the Divergence Theorem states:
\[ \text{If } S \text{ is a closed surface enclosing a volume } V, \text{ then } otag \ \iint_{S} \textbf{V} \bullet \textbf{n} \, d \sigma = \iiint_{V} abla \bullet \textbf{V} \, dV. otag \ otag \ otag \ otag \]In this formula:

• \textbf{V} is a vector field.
• \textbf{n} is the outward-pointing unit normal vector on the surface S.
• abla \bullet \textbf{V} is the divergence of \textbf{V}.
• d \sigma represents an infinitesimal area on the surface S.

This theorem is very useful in simplifying complex integrals, especially when dealing with symmetrical objects, like a sphere. Using the Divergence Theorem, one can often transform a challenging surface integral into a more tractable volume integral.

Vector Field

A vector field assigns a vector to every point in a space. In this exercise, the vector field \textbf{V} is given by:
\[ x \, \cos^{2} y \textbf{i} + xz \textbf{j} + z \, \sin^{2} y \textbf{k}. \ \ otag \]This means at each point (x, y, z), the vector \textbf{V} points in a direction given by the weighted combination of unit vectors i, j, and k.
The components of this vector field are dependent on the coordinates:

• The \textbf{i} component is x \cos^{2} y .
• The \textbf{j} component is xz .
• The \textbf{k} component is z \, \sin^{2} y .
To solve problems involving vector fields, it's important to break down these components and understand how they interact with the surfaces and volumes in question.
In our case, finding the divergence means we compute the partial derivatives of these components relative to their corresponding variables and sum them up: \[ abla \bullet \textbf{V} = \frac{\partial}{ \partial x}(x \, \cos^{2} y ) + \frac{\partial}{ \partial y} (xz) + \frac{\partial }{ \partial z}(z \, \sin^{2} y)= 1.

Volume of a Sphere

In our solution, we converted a surface integral into a volume integral using the Divergence Theorem. To complete the evaluation, we needed to find the volume of a sphere.
The volume of a sphere with radius r is determined by the formula:
\[ V=\frac{4}{3} \pi r^{3}\]In our case, the radius of the sphere is 3. Plugging this into the volume formula gives:\[\frac{4}{3} \pi (3)^3 = \frac{4}{3} \pi 27 = 36 \pi,. \]Understanding the geometry of the problem helped simplify our calculations. Instead of directly evaluating a potentially complicated surface integral over the sphere, we leveraged the symmetry and geometric properties of the sphere, along with the Divergence Theorem, to find that the surface integral is simply \ 36 \pi.\ This underscores an important point in vector calculus: thinking geometrically and using symmetry can vastly simplify solutions.

Volume of a Sphere

In our solution, we converted a surface integral into a volume integral using the Divergence Theorem. To complete the evaluation, we needed to find the volume of a sphere.
The volume of a sphere with radius r is determined by the formula:
\[ V=\frac{4}{3} \pi r^{3}\]In our case, the radius of the sphere is 3. Plugging this into the volume formula gives:\[\frac{4}{3} \pi (3)^3 = \frac{4}{3} \pi 27 = 36 \pi,. \]Understanding the geometry of the problem helped simplify our calculations. Instead of directly evaluating a potentially complicated surface integral over the sphere, we leveraged the symmetry and geometric properties of the sphere, along with the Divergence Theorem, to find that the surface integral is simply \ 36 \pi.\ This underscores an important point in vector calculus: thinking geometrically and using symmetry can vastly simplify solutions.