Problem 388 - 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. Use the divergence theorem to calculate surface integral \(\quad \iint_{S} \mathbf{F} \cdot d \mathbf{S}\) when \(\mathbf{F}(x, y, z)=z \tan ^{-1}\left(y^{2}\right) \mathbf{i}+z^{3} \ln \left(x^{2}+1\right) \mathbf{j}+z \mathbf{k}\) and \(S\) is a part of paraboloid \(x^{2}+y^{2}+z=2\) that lies above plane \(z=1\) and is oriented upward.

Step-by-Step Solution

Answer

The integral evaluates to \(\frac{\pi}{2}\).

Step 1: Recognize the Divergence Theorem

The divergence theorem states that \(\iint_{S} \mathbf{F} \cdot d\mathbf{S} = \iiint_{V} abla \cdot \mathbf{F} \, dV\), where \(V\) is the volume enclosed by the surface \(S\). This theorem allows us to convert a surface integral into a volume integral.

Step 2: Compute the Divergence

First, find the divergence \(abla \cdot \mathbf{F}\) where \(\mathbf{F} = z \tan^{-1}(y^2) \mathbf{i} + z^3 \ln(x^2+1) \mathbf{j} + z\mathbf{k}\). The divergence is given by:\[abla \cdot \mathbf{F} = \frac{\partial}{\partial x}(z \tan^{-1}(y^2)) + \frac{\partial}{\partial y}(z^3 \ln(x^2+1)) + \frac{\partial}{\partial z}(z)\]This simplifies to:\[0 + 0 + 1 = 1\]

Step 3: Define the Volume Integral

Set up the volume integral: \(\iiint_{V} 1 \, dV\), which simplifies to computing the volume of the region bounded by \(x^2 + y^2 + z = 2\) above the plane \(z=1\).The limits for \(z\) are from 1 to \(2 - x^2 - y^2\), and the projection onto the \(xy\)-plane is a circle defined by \(x^2 + y^2 \leq 1\).

Step 4: Convert to Polar Coordinates

Convert the circular region to polar coordinates with \(x = r\cos(\theta)\) and \(y = r\sin(\theta)\).For \(r\), integrate from 0 to 1, and for \(\theta\), integrate from 0 to \(2\pi\).The integral becomes:\[\int_{0}^{2\pi}\int_{0}^{1} \int_{1}^{2-r^2} 1 \cdot z \, dz \, r \, dr \, d\theta\]

Step 5: Evaluate the Volume Integral

First, evaluate the integral with respect to \(z\):\[\int_{1}^{2-r^2} 1 \, dz = [z]_{1}^{2-r^2} = (2 - r^2) - 1 = 1 - r^2\]Next, integrate with respect to \(r\):\[\int_{0}^{1} (1 - r^2) r \, dr = \int_{0}^{1} (r - r^3) \, dr\]Evaluate:\[\left[\frac{r^2}{2} - \frac{r^4}{4}\right]_{0}^{1} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}\]Finally, integrate with respect to \(\theta\):\[\int_{0}^{2\pi} \frac{1}{4} \, d\theta = \frac{1}{4} [\theta]_{0}^{2\pi} = \frac{1}{4}(2\pi) = \frac{\pi}{2}\]

Step 6: Conclusion

The value of the surface integral \(\iint_{S} \mathbf{F} \cdot d \mathbf{S}\) is equal to the volume integral we've computed, which is \(\frac{\pi}{2}\).

Key Concepts

Surface Integral

A surface integral extends the concept of an integral to functions defined on surfaces in three-dimensional space. It is essentially the flux of a vector field through a surface. If you imagine a fluid flowing through a net, the surface integral measures the quantity of the fluid passing through a section of the net.

The mathematical formulation involves integrating the dot product of a vector field \(abla \) with the outward facing unit normal vector \( extbf{n}\) of the surface, over the entire surface area, represented by \( extbf{FW \cdot \textbf{n} \, \,ds}\). For closed surfaces, this gives the total flow or flux through the surface.

When complex vector fields are involved, simplified methods like the divergence theorem can transform surface integrals into volume integrals, making calculations more manageable.

Volume Integral

A volume integral calculates the integral of a function over a three-dimensional region. Within the context of the divergence theorem, it replaces the surface integral and allows for a simpler evaluation of the flux. The essence of the theorem is:

• The surface integral of a field \(\textbf{F}\) over a closed surface \(S\) equals the volume integral of the divergence of \(\textbf{F}\) throughout the volume \(V\) bounded by \(S\).
• This is expressed as \(abla \cdot extbf{F} \, dV\), where \(abla \cdot extbf{F}\) is the divergence.

In our solution, the volume integral \(\iiint_{V} 1 \, dV\) was calculated, representing the space over which the surface \(S\) extends. This helped to transform and solve the original surface integral problem into a more straightforward calculation of volume.

Polar Coordinates

Polar coordinates are useful for converting integrals over circular regions from Cartesian coordinates (x, y) to a system better suited for circular symmetry.

In polar coordinates:

• The position of a point in a plane is determined by a distance from a reference point (the origin) and an angle from a reference direction.
• The conversion is defined by: \(x = r \cos(\theta)\), \(y = r \sin(\theta)\), where \(r\) is the radius and \(\theta\) the angle.

In the exercise, polar coordinates simplified the integration over the circular surface \(x^2 + y^2 \leq 1\), as the variables naturally fit the region's geometry. This transformation makes both setting up and evaluating the integral far more efficient.

Computer Algebra System

A Computer Algebra System (CAS) is a software tool that facilitates symbolic mathematics. It handles tasks such as integration, differentiation, and solving algebraic equations with precision and efficiency.

Using a CAS can simplify finding the divergence or solving integrals in the exercises by:

• Automating tedious algebraic calculations, allowing for faster and error-free solutions.
• Enabling visualization tools to better understand the geometric nature of problems.
• Providing step-by-step solutions for educational purposes, helping solidify understanding of complex concepts.

In the complex problem-solving environment of calculus and vector fields, a CAS proves invaluable for executing intricate steps and verifying computations that adhere closely to theoretical frameworks, like the divergence theorem.