Question

Evaluate each integral in the simplest way possible.\(\iint \mathbf{F} \cdot \mathbf{n} d \sigma\) where \(\mathbf{F}=\left(y^{2}-x^{2}\right) \mathrm{i}+(2 x y-y) \mathrm{j}+3 z \mathbf{k}\) and \(\sigma\) is the entire surfacoce of the tin can bounded by the cylinder \(x^{2}+y^{2}=16, z=3, z=-3\).

Short answer

192\pi

Step-by-step solution

Understand the Problem

We need to evaluate the surface integral \(\iint \mathbf{F} \cdot \mathbf{n} \, d \sigma\) where \(\mathbf{F} = (y^{2}-x^{2}) \mathbf{i} + (2xy-y) \mathbf{j} + 3z \mathbf{k}\). The surface \(\sigma\) is the entire surface of the tin can described by the cylinder \(x^{2}+y^{2}=16, z=3, z=-3\).

Apply the Divergence Theorem

The Divergence Theorem states \[\iint_{\sigma} \mathbf{F} \cdot \mathbf{n} \, d \sigma = \iiint_{V} abla \cdot \mathbf{F} \, dV\]. Therefore, we need to find the volume integral of the divergence of \(\mathbf{F}\) over the volume \(V\) enclosed by the surface.

Compute the Divergence of \(\mathbf{F}\)

To find abla \cdot \mathbf{F}, we compute the partial derivatives and sum them: \[abla \cdot \mathbf{F} = \frac{\partial}{\partial x}(y^{2}-x^{2}) + \frac{\partial}{\partial y}(2xy-y) + \frac{\partial}{\partial z}(3z)\]. Here, \[\frac{\partial}{\partial x}(y^{2}-x^{2}) = -2x,\] \[\frac{\partial}{\partial y}(2xy-y) = 2x-1,\] \[\frac{\partial}{\partial z}(3z) = 3\]. Therefore, \[abla \cdot \mathbf{F} = -2x + (2x - 1) + 3 = 2.\]

Setup and Evaluate the Volume Integral

The volume \(V\) is the region inside the cylinder \(x^{2}+y^{2}=16\) between \(z=-3\) and \(z=3\). Using cylindrical coordinates (where \(x=r\cos\theta\) and \(y=r\sin\theta\)) simplifies the integral: \[\iiint_{V} 2 \, dV = 2 \iiint_{V} \, dV.\] The limits are \(0 \leq r \leq 4\), \(0 \leq \theta \leq 2\pi\), and \(-3 \leq z \leq 3\). Thus, \[\iiint_{V} \, dV = \int_{0}^{2\pi} \int_{0}^{4} \int_{-3}^{3} r \, dz \, dr \, d\theta.\] Evaluating the integrals step-by-step: \[\begin{aligned} &\int_{-3}^{3} dz = 6,\ &\int_{0}^{4} r \, dr = \frac{r^{2}}{2}\big|_{0}^{4} = 8,\ &\int_{0}^{2\pi} d\theta = 2\pi. \end{aligned}\] Combining these results, \[\begin{aligned} \iiint_{V} \, dV &= 6 \cdot 8 \cdot 2\pi = 96\pi. \end{aligned}\] Therefore, \[\iiint_{V} abla \cdot \mathbf{F} \, dV = 2 \cdot 96\pi = 192\pi.\]

Key concepts

Divergence Theorem

The Divergence Theorem is a powerful tool in vector calculus that relates a flux integral over a closed surface to a volume integral of the divergence over the region enclosed by the surface. This theorem simplifies the problem significantly if the conditions are met. Mathematically, the Divergence Theorem states: \[\iint_{\sigma} \mathbf{F} \cdot \mathbf{n} \, d \sigma = \iiint_{V} abla \cdot \mathbf{F} \, dV \]Here, \(\mathbf{F}\) is a vector field, \(\mathbf{n}\) is the outward-pointing unit normal vector on the surface \(\sigma\), and \(V\) is the volume enclosed by \(\sigma\). When applying the Divergence Theorem, follow these steps:

• Compute the divergence of the provided vector field, \(\mathbf{F}\).
• Set up the volume integral over the region \(V\) enclosed by the surface.
• Integrate to find the total flux through the surface.

It greatly simplifies many problems involving flux integrals, especially when working with symmetrical regions like cylinders or spheres.

Cylindrical Coordinates

Cylindrical coordinates are useful when dealing with problems that have symmetry about an axis, such as cylinders and cones. In Cartesian coordinates, a point is represented as \(x, y, z\), while in cylindrical coordinates, it is represented as \(r, \theta, z\). The transformations between these two systems are:\[x = r \cos \theta,\quad y = r \sin \theta,\quad \text{and}\quad z = z. \]Using cylindrical coordinates can simplify the integration process for regions with circular cross-sections. For instance, to integrate a volume integral in cylindrical coordinates, the differential element of volume changes from \(dV = dx\, dy\, dz\) to \(dV = r\, dr\, d\theta\, dz\).

• \(r\) is the radius from the z-axis to the point in the xy-plane (non-negative).
• \(\theta\) is the angle from the positive x-axis to the radius in the xy-plane.
• \(z\) remains the same in both coordinate systems.

In this exercise, we used cylindrical coordinates to evaluate the volume integral inside the cylinder \(x^2 + y^2 = 16\) bounded by \(z = -3\) and \(z=3\). This conversion helps manage the integral's limits more easily.

Vector Fields

A vector field assigns a vector to every point in space. It can represent various physical quantities, such as velocity fields in fluid dynamics or electromagnetic fields. In this exercise, the given vector field \(\mathbf{F}\) is expressed as \(\mathbf{F} = (y^2 - x^2) \mathbf{i} + (2xy - y) \mathbf{j} + 3z \mathbf{k}\). Each component function corresponds to one of the standard basis vectors, \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\).To evaluate integrals involving vector fields, we often need to compute their divergence or curl. For a vector field \(\mathbf{F} = (F_1, F_2, F_3)\), the divergence is computed as:\[abla \cdot \mathbf{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z} \]In this problem, computing the divergence \(abla \cdot \mathbf{F}\) involves taking partial derivatives of each component and summing those derivatives. This operation transforms a vector field into a scalar field.

Multivariable Calculus

Multivariable calculus extends the principles of calculus to functions of multiple variables. Key topics include partial derivatives, multiple integrals, and vector calculus. In this exercise, we utilized several concepts from multivariable calculus:

• Partial Derivatives: Calculating derivatives of functions with respect to one variable while holding others constant.
• Volume Integrals: Integrating functions over three-dimensional regions, which can be simplified using cylindrical coordinates.
• Divergence: A measure of a vector field's tendency to originate from or converge at points in space — critical for applying the Divergence Theorem.

By mastering multivariable calculus, students can tackle complex problems in physics, engineering, and other fields involving multiple dimensions. This includes handling applications settings like electromagnetic fields, fluid flow, and more. Understanding these foundational concepts is essential for solving real-world problems efficiently.