Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. The work required to move an object around a closed curve \(C\) in the presence of a vector force field is the circulation of the force field on the curve. b. If a vector field has zero divergence throughout a region (on which the conditions of Green's Theorem are met), then the circulation on the boundary of that region is zero. c. If the two-dimensional curl of a vector field is positive throughout a region (on which the conditions of Green's Theorem are met), then the circulation on the boundary of that region is positive (assuming counterclockwise orientation).

Short answer

Question: Determine if the following statements are true or false and provide explanations or counterexamples. a) The double integral of a function on a flat surface S is equal to the given integral: \(\iint_{S} f(x, y, 10) d S= \int_{0}^{1} \int_{0}^{1} f(x, y, 10) d x d y\). b) The double integral of a function on the sloping surface S is equal to the given integral: \(\iint_{S} f(x, y, x) dS = \int_{0}^{1} \int_{0}^{1} f(x, y, x) dx dy\). c) The parameterizations \(\mathbf{r}_1(v, u) = \left\langle v \cos u, v \sin u, v^{2}\right\rangle\) and \(\mathbf{r}_2(v, u) = \left\langle\sqrt{v} \cos 2u, \sqrt{v} \sin 2u, v\right\rangle\) represent the same surface. d) The normal vectors on a standard parameterization of a sphere are outward.

Step-by-step solution

a. Flat Surface BoxFit

Given the surface S, we can see that it is a flat surface with z = 10. In this case, we do not have to adjust the integrand for the surface integral. Therefore, the given surface integral is equal to the double integral: \(\iint_{S} f(x, y, 10) d S= \int_{0}^{1} \int_{0}^{1} f(x, y, 10) d x d y\) So, the statement is true.

b. Sloping Surface BoxFit

For the surface S, defined by z = x, we need to find the surface integral. To do this, we first need to find the surface element dS. Using a regular Riemann sum, dS = sqrt(1 + (dz/dx)^2 + (dz/dy)^2) dx dy. Since the equation z = x has the partial derivatives dz/dx = 1 and dz/dy = 0, dS = sqrt(2) dx dy. The surface integral becomes: \(\iint_{S} f(x, y, x) dS =\iint_{R} f(x, y, x)\sqrt{2} dx dy\), where R is the domain of x and y, and in this case, R = {[0,1]x[0,1]}. However, the given integral is: \(\int_{0}^{1} \int_{0}^{1} f(x, y, x) dx dy\) Therefore, the statement is false.

c. Surface Comparison

We are given two parameterizations of a surface: \(\mathbf{r}_1(v, u) = \left\langle v \cos u, v \sin u, v^{2}\right\rangle\) with \(0 \leq u \leq \pi\) and \(0 \leq v \leq 2\) \(\mathbf{r}_2(v, u) = \left\langle\sqrt{v} \cos 2u, \sqrt{v} \sin 2u, v\right\rangle\) with \(0 \leq u \leq \frac{\pi}{2}\) and \(0 \leq v \leq 4\) To compare these two surfaces, we can observe how they change with respect to u and v or the shapes formed by them. Analyzing both parameterizations, we can see that they are not equivalent, and they generate different surfaces. The second parameterization generates a different shape due to the square roots and 2u angles compared to the first parameterization. Therefore, the statement is false.

d. Sphere Normal Vector

Let's consider the standard parameterization of a sphere of radius R: \(\mathbf{r}(u, v) = \left\langle R\sin v\cos u, R\sin v\sin u, R\cos v\right\rangle\) with \(0 \leq u \leq 2\pi\) and \(0 \leq v \leq \pi\). To find the tangent vectors, we take the partial derivatives with respect to u and v: \(\mathbf{t}_u = \frac{\partial\mathbf{r}}{\partial u} = \left\langle -R\sin v\sin u, R\sin v\cos u, 0\right\rangle\) \(\mathbf{t}_v = \frac{\partial\mathbf{r}}{\partial v} = \left\langle R\cos v\cos u, R\cos v\sin u, -R\sin v\right\rangle\) Now we calculate the cross product of these vectors to find the normal vector: \(\mathbf{n}=\mathbf{t}_u\times\mathbf{t}_v=\left\langle R^2\sin^2v\cos u, R^2\sin^2v\sin u, R^2\sin v\cos v\right\rangle\) Considering the direction of each component, we can see that the normal vectors point outward from the sphere, being in the direction of the position vector. Thus, the statement is true.

Key concepts

Double Integral

Double integrals are a form of integration used to calculate the volume under a surface in a two-dimensional space. In essence, the calculation involves adding up infinitesimally small areas over a region to obtain the total quantity of interest. The notation for a double integral is expressed as \(\iint_D f(x, y) \, dx \, dy\), where \(f(x, y)\) is the function defined over the domain \(D\), and \(dx\) and \(ddy\) represent the infinitesimal increments in the x and y directions, respectively.

Think of it as slicing the region into tiny rectangles, each with an area of \(dx \cdot dy\), and then summing up the volume slices, which is done by multiplying the area of each rectangle by the value of the function at that point. This process totals the volumes across the region to yield the desired integral value.

Parameterization of a Surface

The process of parameterization involves representing a surface in terms of two independent parameters, often denoted \(u\) and \(v\). The main idea is to express the coordinates \(x(u, v)\), \(y(u, v)\), and \(z(u, v)\) of points on the surface as functions of these parameters. This description allows us to move from the Cartesian coordinates to working with the surface in terms of these parameters, which can be highly useful when performing tasks such as integration over the surface.

Parameterization is particularly powerful because it enables us to capture complex surfaces through simpler mathematical representations. For instance, to evaluate a surface integral where a directly suitable x-y domain isn't apparent, we can parameterize the surface, switching the challenging integration to be over a typically more manageable parameter domain.

Normal Vectors

Normal vectors are perpendicular to a surface at a given point, and they are fundamental in calculating surface integrals and in vector calculus as a whole. Finding the normal vector often involves computing the cross product of two tangent vectors to the surface, which are derived from the partial derivatives with respect to the parameters used in the surface's parameterization.

In the context of surface integration, the normal vector helps in defining the orientation of the surface, which in turn helps us understand the direction in which we're 'accumulating' the quantities. For instance, whether the fluid through a surface is exiting or entering a volume, or whether the electric field is entering or leaving a closed surface, depends on the direction of the normal vectors.

Sphere Parameterization

Sphere parameterization involves describing the coordinates of points on the surface of a sphere using two parameters, typically azimuthal and polar angles denoted by \(u\) and \(v\), respectively. As showcased in the solution concerning sphere normalization, the standard parameterization of a sphere with radius \(R\) is \(\mathbf{r}(u, v) = \left\langle R\sin v\cos u, R\sin v\sin u, R\cos v\right\rangle\), where \(0 \leq u \leq 2\pi\) and \(0 \leq v \leq \pi\).

This parameterization is instrumental in computations involving spheres, such as calculating the surface area, surface integrals, or other geometric properties. Moreover, understanding this parameterization helps visualize and work with the sphere in a more mathematically manageable way, particularly when dealing with physics problems or geometric considerations where the sphere’s symmetry simplifies calculations.