Problem 382 - Calculus Volume 3

Question

For the following exercises, use a computer algebraic system (CAS) and the divergence theorem to evaluate surface integral \(\int_{S} \mathbf{F} \cdot \mathbf{n} d s\) for the given choice of \(\mathbf{F}\) and the boundary surface \(S\). For each closed surface, assume \(\mathbf{N}\) is the outward unit normal vector. [T] \(\mathbf{F}(x, y, z)=x y^{2} \mathbf{i}+y z^{2} \mathbf{j}+x^{2} z \mathbf{k} ; \quad S\) is the surface bounded above by sphere \(\rho=2\) and below by cone \(\varphi=\frac{\pi}{4}\) in spherical coordinates. (Think of \(S\) as the surface of an "ice cream cone.")

Step-by-Step Solution

Answer

The surface integral evaluates to \(\frac{64\pi\sqrt{2}}{5}\).

Step 1: Understanding the Problem

We are given a vector field \(\mathbf{F}(x, y, z)=x y^{2} \mathbf{i}+y z^{2} \mathbf{j}+x^{2} z \mathbf{k}\) and a closed surface \(S\) which is bounded above by the sphere \(\rho=2\) and below by the cone \(\varphi=\frac{\pi}{4}\). We'll evaluate the surface integral \(\int_{S} \mathbf{F} \cdot \mathbf{n} \, dS\) using the Divergence Theorem.

Step 2: Apply the Divergence Theorem

The Divergence Theorem states that for a vector field \(\mathbf{F}\) and a closed surface \(S\), \(\int_{S} \mathbf{F} \cdot \mathbf{n} \, dS = \int_{V} abla \cdot \mathbf{F} \, dV\), where \(V\) is the volume enclosed by \(S\) and \(abla \cdot \mathbf{F}\) is the divergence of \(\mathbf{F}\).

Step 3: Calculate the Divergence

Find the divergence of \(\mathbf{F}\), which is \(abla \cdot \mathbf{F} = \frac{\partial}{\partial x}(xy^2) + \frac{\partial}{\partial y}(yz^2) + \frac{\partial}{\partial z}(x^2z) = y^2 + z^2 + x^2\).

Step 4: Set Up the Triple Integral

Convert the volume integral into spherical coordinates since the boundary includes a sphere and a cone. The divergence becomes \(y^2 + z^2 + x^2 = \rho^2\). We need to evaluate \(\int_{V} \rho^2 \, dV\) over the region bounded by \(\rho=2\) and the cone \(\varphi=\frac{\pi}{4}\).

Step 5: Spherical Coordinates Bounds

The bounds for spherical coordinates are: \(0 \leq \theta < 2\pi\), \( \frac{\pi}{4} \leq \varphi \leq \frac{\pi}{2} \), and \(0 \leq \rho \leq 2\).

Step 6: Evaluate the Integral

We calculate the integral \(\int_0^{2\pi}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\int_0^2 \rho^2 \rho^2 \sin \varphi \, d\rho \, d\varphi \, d\theta\), because \(dV = \rho^2 \sin \varphi \, d\rho \, d\varphi \, d\theta\).

Step 7: Integral Computation

Perform the integration step-by-step. First, integrating with respect to \(\rho\) from 0 to 2 gives \(\frac{\rho^5}{5}\bigg|_0^2 = \frac{32}{5}\). Then, integrate with respect to \(\varphi\) from \(\frac{\pi}{4}\) to \(\frac{\pi}{2}\), getting the integral of \(\sin \varphi\) which results in \(-\cos \varphi\bigg|_{\frac{\pi}{4}}^{\frac{\pi}{2}} = -(-\frac{\sqrt{2}}{2}) = \frac{\sqrt{2}}{2}\). Finally, integrate with respect to \(\theta\) from 0 to \(2\pi\): \(\int_0^{2\pi} d\theta = 2\pi\).

Step 8: Final Result

Combine all parts: \(\frac{32}{5} \times \frac{\sqrt{2}}{2} \times 2\pi = \frac{64\pi\sqrt{2}}{5}\). Thus, the value of the surface integral is \(\frac{64\pi\sqrt{2}}{5}\).

Key Concepts

Surface Integrals

Surface integrals extend the concept of a definite integral to summing over a surface in three-dimensional space. Instead of summing slices, as you would with a regular integral, a surface integral adds up quantities over a surface, which can be curved or flat. Consider a surface, like the top of a scoop of ice cream, and a vector field, such as a wind flow or electromagnetic field, defined over that surface.

• The surface integral \(\int_{S} \mathbf{F} \cdot \mathbf{n} \, dS\) computes how much the vector field \(\mathbf{F}\) "flows" through the surface \((S)\).
• Here, \(\mathbf{n}\) is the unit normal vector pointing outward from \(S\).
• Imagine shining a set of parallel light rays onto the surface; the surface integral gives a sense of how much light would "pass" through, factoring in angle and strength of the field.

When using the Divergence Theorem, you can convert a surface integral into a volume integral, often simplifying calculations considerably. This is particularly handy in physics and engineering when assessing flux through gases or electromagnetic fields.

Spherical Coordinates

Spherical coordinates provide a natural way to represent points in 3D space using three variables: radial distance, angle from the polar axis, and azimuthal angle. It's useful when dealing with problems involving spheres, such as celestial bodies like planets.

• \(\rho\) is the radial distance from the origin to the point. Think of it as the length of a straight line from the center of a sphere to its edge.
• \(\varphi\) is known as the polar angle, which is measured from the positive z-axis. It's literally the angle you make from the top downwards (like cutting a pizza, but vertically).
• \(\theta\) is the azimuthal angle in the xy-plane, measured from the positive x-axis. This is like the hands of a clock moving around the face.

This coordinate system is particularly useful for integrating over regions bounded by spherical surfaces, like our ice cream cone scenario. Each coordinate handles a different aspect (distance, height, sweep), allowing comprehensive coverage of 3D space. Converting to spherical coordinates makes solving integrals over radial symmetries straightforward and efficient.

Vector Field Divergence

Divergence measures the "outflowing-ness" of a vector field from an infinitesimally small region. It's a key concept in vector calculus, providing insight into behavior of flows. Imagine it as quantifying how much a vector field spreads out or converges.

• Given a vector field \(\mathbf{F}(x, y, z) = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k}\), the divergence is computed as \(abla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}\).
• If \(abla \cdot \mathbf{F} > 0\), the field is generally diverging (like air from a balloon).
• If \(abla \cdot \mathbf{F} < 0\), it's converging or compressing.

In the original problem, we calculated the divergence of \(\mathbf{F}(x, y, z) = xy^2\mathbf{i} + yz^2\mathbf{j} + x^2z\mathbf{k}\) to be \(y^2 + z^2 + x^2\). This calculation forms the foundation for switching from a surface integral to a volume integral over the encompassed region using the Divergence Theorem. Understanding divergence helps to simplify complex interactions within vector fields by focusing on scalar quantities, making computations easier.