Problem 390 - Calculus Volume 3

Question

Use the divergence theorem to compute the value of flux integral \(\iint_{S} \mathbf{F} \cdot d \mathbf{S}\), where \(\mathbf{F}(x, y, z)=\left(y^{3}+3 x\right) \mathbf{i}+(x z+y) \mathbf{j}+\left[z+x^{4} \cos \left(x^{2} y\right) \mathbf{k}\right.\) and \(S\) is the area of the region bounded by \(x^{2}+y^{2}=1, x \geq 0, y \geq 0,\) and \(0 \leq z \leq 1\)

Step-by-Step Solution

Answer

The flux integral is \(\frac{5\pi}{4}\).

Step 1: Define the Volume and the Vector Field

Identify the region of space for which the flux needs to be calculated. The volume defined here is the solid bounded by the circular region in the xy-plane given by \(x^2 + y^2 = 1\) in the first quadrant \((x \geq 0, y \geq 0)\) and extending from \(z = 0\) to \(z = 1\). The vector field \(\mathbf{F}\) is given by \(\mathbf{F}(x, y, z) = (y^3 + 3x)\mathbf{i} + (xz + y)\mathbf{j} + \left(z + x^4 \cos(x^2y)\right)\mathbf{k}\).

Step 2: State the Divergence Theorem

Apply the Divergence Theorem, which states that the flux integral over a closed surface \(S\) of a vector field \(\mathbf{F}\) is equal to the triple integral of the divergence of \(\mathbf{F}\) over the volume \(V\) enclosed by \(S\): \( \iint_{S} \mathbf{F} \cdot d\mathbf{S} = \iiint_{V} abla \cdot \mathbf{F} \, dV \).

Step 3: Calculate the Divergence of \(\mathbf{F}\)

Calculate \(abla \cdot \mathbf{F}\). The divergence of \(\mathbf{F}(x, y, z)\) is computed as: \(\frac{\partial}{\partial x}(y^3 + 3x) + \frac{\partial}{\partial y}(xz + y) + \frac{\partial}{\partial z}(z + x^4 \cos(x^2 y))\). Calculate each partial derivative: - \(\frac{\partial}{\partial x}(y^3 + 3x) = 3\)- \(\frac{\partial}{\partial y}(xz + y) = 1\)- \(\frac{\partial}{\partial z}(z + x^4 \cos(x^2 y)) = 1\)Combine these to find \(abla \cdot \mathbf{F} = 3 + 1 + 1 = 5\).

Step 4: Set Up the Triple Integral

The triple integral becomes \(\iiint_{V} 5 \, dV\). Since \(abla \cdot \mathbf{F} = 5\), and the region \(V\) is cleverly chosen, we consider the volume as the cylindrical portion restricted to the first quadrant: \(0 \leq x \leq 1\), \(0 \leq y \leq \sqrt{1-x^2}\), and \(0 \leq z \leq 1\).

Step 5: Evaluate the Integral

Calculate the triple integral in cylindrical coordinates. Consider the symmetry and the limits of integration:\[ \int_{z=0}^{1} \int_{y=0}^{\sqrt{1-x^2}} \int_{x=0}^{1} 5 \, dx \, dy \, dz \]This is easiest with cylindrical coordinates: \(r \) from \(0\) to \(1\), \(\theta\) from \(0\) to \(\frac{\pi}{2}\), and \(z\) from \(0\) to \(1\):\[ \int_{z=0}^{1} \int_{\theta=0}^{\pi/2} \int_{r=0}^{1} 5r \, dr \, d\theta \, dz \]

Step 6: Solve the Integral

Complete the integral step-by-step:- Integral with respect to \(r\): \(\int_{r=0}^{1} 5r \, dr = 5\left[ \frac{r^2}{2} \right]_{0}^{1} = \frac{5}{2}\)- Integral with respect to \(\theta\): \(\int_{\theta=0}^{\pi/2} \frac{5}{2} \, d\theta = \frac{5}{2} \cdot \frac{\pi}{2} = \frac{5\pi}{4}\)- Integral with respect to \(z\): \(\int_{z=0}^{1} \frac{5\pi}{4} \, dz = \frac{5\pi}{4}\).The total flux integral evaluates to \(\frac{5\pi}{4}\).

Key Concepts

Flux Integral

A flux integral essentially measures the flow of a vector field through a surface. If you imagine a fluid moving through space, the flux is the amount of fluid passing through a given surface per unit time. It is a vital concept in fields like physics and engineering, helping in understanding fluid dynamics and electromagnetism.
To calculate the flux integral, we often use the surface integral of a vector field. Mathematically, it's expressed as \(\iint_{S} \mathbf{F} \cdot d\mathbf{S}\), where \(\mathbf{F}\) is the vector field, and \(d\mathbf{S}\) is a small element of the area \(S\) through which the field flows.
In practice, calculating the flux integral directly can be quite complex, especially for irregularly shaped surfaces. However, the Divergence Theorem provides a handy shortcut by converting a surface integral into a volume integral, simplifying the computation significantly.

Vector Field

A vector field assigns a vector to every point in space, providing a way to model the direction and magnitude of a physical quantity. For example, gravitational forces, magnetic fields, and velocity fields in fluid flow can all be described using vector fields.
Given in the form \(\mathbf{F}(x, y, z) = (P(x, y, z), Q(x, y, z), R(x, y, z))\), each component \(P, Q, \text{and} R\) might depend on the spatial coordinates \(x, y,\) and \(z\). This representation helps in visualizing how a field behaves at different spatial points.
In our exercise, the vector field \(\mathbf{F}(x, y, z) = (y^3+3x, xz+y, z+x^4 \cos(x^2y))\) dictates how the field aligns throughout the region defined by \(S\). By examining the divergence of this vector field, we can determine the outflow through the surface \(S\) using the Divergence Theorem.

Triple Integral

Triple integrals allow us to integrate over three-dimensional regions, providing a powerful tool for calculating volumes, mass, and other quantities in 3D space. Particularly useful when applying the Divergence Theorem, it lets us transform the complex calculation of surface flux into a simpler problem of volume integration.
For a function \(f(x, y, z)\) defined in a region \(V\), the triple integral is expressed as \(\iiint_{V} f(x, y, z) \, dV\). Here, \(dV\) is a small volume element, which can be aligned with the coordinate system: Cartesian, cylindrical, or spherical coordinates based on the problem's symmetry.
In this exercise, the triple integral \(\iiint_{V} 5 \, dV\) helps us compute the flux through the region. The divergence \(abla \cdot \mathbf{F} = 5\) indicates constant outflow across the volume \(V\), simplifying the triple integral calculation as it avoids complex expressions in the integrand.

Cylindrical Coordinates

Cylindrical coordinates offer a bridge between Cartesian coordinates and polar coordinates, suitable for problems possessing rotational symmetry around an axis. Defined as \((r, \theta, z)\) — where \(r\) is the radius, \(\theta\) the angle in the xy-plane, and \(z\) the height — they simplify integration over circular or cylindrical regions.
These coordinates are particularly handy when the volume or region you want to integrate over has a circular base, like cylinders or cones. Here, changing variables from Cartesian to cylindrical coordinates cuts down the complexity of the integration, as seen in our exercise.
For our problem, the integral in cylindrical coordinates becomes \(\int_{z=0}^{1} \int_{\theta=0}^{\pi/2} \int_{r=0}^{1} 5r \, dr \, d\theta \, dz\). This representation showcases how radial symmetry helps break down the integral into simpler, nested forms across varying dimension ranges — starting from the radius, then angle, and finishing with height.