Question

Determine whether the following statements are true and give an explanation or counterexample. a. If the surface \(S\) is given by \(\\{(x, y, z): 0 \leq x \leq 1,0 \leq y \leq 1\) \(z=10\\},\) then \(\iint_{S} f(x, y, z) d S=\int_{0}^{1} \int_{0}^{1} f(x, y, 10) d x d y\) b. If the surface \(S\) is given by \(\\{(x, y, z): 0 \leq x \leq 1,0 \leq y \leq 1\) \(z=x\\},\) then \(\iint_{S} f(x, y, z) d S=\int_{0}^{1} \int_{0}^{1} f(x, y, x) d x d y\) c. The surface \(\mathbf{r}=\left\langle v \cos u, v \sin u, v^{2}\right\rangle,\) for \(0 \leq u \leq \pi\) \(0 \leq v \leq 2,\) is the same as the surface \(\mathbf{r}=\langle\sqrt{v} \cos 2 u, \sqrt{v} \sin 2 u, v\rangle,\) for \(0 \leq u \leq \pi / 2,0 \leq v \leq 4\) d. Given the standard parameterization of a sphere, the normal vectors \(\mathbf{t}_{u} \times \mathbf{t}_{v}\) are outward normal vectors.

Short answer

Answer: True Explanation: Since the surface is a flat plane, the surface integral equals the double integral, so the statement is true.

Step-by-step solution

a. Analyzing the surface and evaluating the surface integral

For this surface S, it's a flat rectangular surface in the 3D space with one corner at the point (0,0,10) and the opposite corner at the point (1,1,10). We can represent this surface as a vector function r(x,y) = . The integral given is a double integral with the limits of x and y defined in the range [0,1]. Since the surface is a flat plane, the surface integral equals the double integral, so the statement is true.

b. Analyzing the surface and evaluating the surface integral

For this surface S, it's a curved shape in 3D space with its boundary defined by x=y=z for x and y in the range of [0,1]. We can represent this surface as a vector function r(x,y) = . The integral given is a double integral with the limits of x and y defined in the range [0,1]. But the surface is not flat, so the surface integral equals the double integral, so the statement is false.

c. Comparing the two surfaces and checking if they are the same

Given two surfaces, we need to compare them to see if they are the same: Surface 1: r(u,v) = with u in [0,pi] and v in [0,2] Surface 2: r(u,v) = with u in [0,pi/2] and v in [0,4] If we test some points for the surfaces, we notice they don't define the same surface. For example, if we use (u,v) = (0,1) for the first surface, we get the point (1,0,1), but for the second surface, we get the point (1,0,1). If we use (u,v) = (pi,1) for the first surface, we get the point (-1, 0, 1), and for the second surface, we get the point (0, 0, 1). Therefore, they don't represent the same surface, and so the statement is false.

d. Checking the normal vectors for the standard sphere parameterization

In this statement, we have a standard parameterization of a sphere: r(u,v) = , with 0 ≤ u ≤ 2*pi and 0 ≤ v ≤ pi To determine the normal vectors, we first calculate the partial derivatives: rt_u(u,v) = <-R*sin(u)*sin(v), R*cos(u)*sin(v), 0> rt_v(u,v) = Now we compute their cross product: rt_u × rt_v = Since R > 0 and sin(v)*cos(v) > 0 for 0 ≤ v ≤ pi, we can conclude that the normal vectors obtained from the cross product rt_u × rt_v are outward normal vectors, and the statement is true.

Key concepts

Double Integral

The concept of a double integral is fundamental when we talk about computing the volume under a surface in a three-dimensional space. When we integrate a function over a two-dimensional region, we essentially sum up all the infinitesimally small areas, each multiplied by the value of the function at that point. In the context of surface integrals, a double integral is responsible for aggregating the values not just over a flat plane, but potentially over curved surfaces as well.

For example in our exercise, for the flat plane surface defined by z = 10, the double integral computes the accumulated effect of the function f(x, y, z) over the rectangular domain x and y between 0 and 1. This is an application where the surface is directly tied to a double integral with no extra considerations for curvature.

Surface Parameterization

Surface parameterization involves the description of a surface using two variables, usually u and v, rather than x, y, and z. This method is particularly useful when dealing with complex surfaces where the direct relation to Cartesian coordinates is not simple. A parameterized surface allows computations, like integrals, to be performed more manageably.

In our second example, where the surface is not a plane, we can't evaluate the integral merely as a double integral. The curvature has to be accounted for, which is where parameterization can assist. If the surface S was parameterized appropriately, we could have reformed the initial integral to respect the shape's geometry.

Normal Vectors

Normal vectors are essential in calculus, specifically in surface integrals, as they represent perpendicular vectors to surfaces at each point. They are crucial for defining orientation and for calculations that involve flux. In our final example involving the sphere, we were interested in finding whether the calculated normal vectors pointed outwards.

To obtain a normal vector to a surface parameterized by a vector function r(u,v), we take the cross product of the partial derivatives r_u and r_v. This gives us a vector that is perpendicular to both r_u and r_v, and hence perpendicular to the surface at the point.

Vector Function

A vector function is a rule that assigns a vector to each point in a domain. In the context of surfaces, vector functions output a vector corresponding to a point on the surface for a given pair of (u,v) parameters. Such functions encapsulate the idea that surfaces can be described in terms of two varying parameters rather than directly through x, y, and z coordinates.

For instance, in our third example, we had two different vector functions supposed to define the same surface. By comparing outputs from these functions, we could determine that they actually defined different surfaces, illustrating the importance of the representation of surfaces by vector functions.

Cross Product

The cross product is an operation on two vectors that returns a third vector which is perpendicular to the plane containing the initial vectors. Its magnitude is equal to the area of the parallelogram that the two vectors span. This concept is particularly relevant when finding normal vectors to surfaces in multivariable calculus.

As seen in part (d) of our exercise, by taking the cross product of r_u and r_v, we don't just get any normal vector; we obtain the one necessary to calculate the outward flux across a surface. The properties of the cross product, such as its direction given by the right-hand rule, underpin many critical calculations in vector calculus.