Question

Determine whether the following statements are true and give an explanation or counterexample. a. An iterated integral of a function over the box \(D=\\{(x, y, z): 0 \leq x \leq a, 0 \leq y \leq b, 0 \leq z \leq c\\}\) can be expressed in eight different ways. b. One possible iterated integral of \(f\) over the prism \(D=\\{(x, y, z): 0 \leq x \leq 1,0 \leq y \leq 3 x-3,0 \leq z \leq 5\\}\) is \(\int_{0}^{3 x-3} \int_{0}^{1} \int_{0}^{5} f(x, y, z) d z d x d y\) \(\begin{aligned} &\text { c. The region } D=\\{(x, y, z): 0 \leq x \leq 1,0 \leq y \leq \sqrt{1-x^{2}}\\\ &0 \leq z \leq \sqrt{1-x^{2}}\\} \text { is a sphere. } \end{aligned}\)

Short answer

a) True. Converting the given integral into polar coordinates results in the same expression. b) False. Since the only points on the hemisphere with a distance equal to 2 from the origin are on the equator, the average distance must be strictly between 0 and 2. c) False. While the integral looks simpler in polar coordinates, evaluating the integral in both coordinate systems is still not straightforward, and neither leads to an elementary function.

Step-by-step solution

Part a - Converting to polar coordinates and compare the integrals

To find out if the given equality holds, we will first transform the given integral in polar coordinates, then compare both expressions. Recall that \(x^2 + y^2 = r^2\), and the disk R can be represented as \(0 \le r \le 1\) and \(0 \le \theta \le 2\pi\). Note that the area differential element \(dA\) in polar coordinates is expressed as \(rd\theta dr\). So we convert the integral as: $\iint_{R}\left(x^{2}+y^{2}\right) d A = \int_{0}^{2 \pi} \int_{0}^{1} r^{2}\cdot r d r d \theta$ The given integral and the transformed one are exactly the same. Therefore, the statement is true.

Part b - Average distance of hemisphere points to the origin

We are given the equation of the hemisphere \(z=\sqrt{4-x^2-y^2}\) and asked to determine if the average distance of these points from the origin is 2. Recall that finding the distance between two points \((x, y, z)\) and \((0, 0, 0)\) in three-dimensional space is calculated by: \(distance = \sqrt{x^2 + y^2 + z^2}\). We are asked to consider the average distance, so we don't need to go through the actual calculation for a specific point. Instead, we can, in this case, inspect the bound on the hemisphere equation: \(z = \sqrt{4-x^2-y^2}\). The smallest value \(z\) can take is 0, and the largest is 2; this means that distances from the origin vary between 0 and 2 (or a sphere of radius 2 in this case). However, if we restrict it to the hemisphere, we see there's no point on the hemisphere with a distance of 0 from the origin, and the closest points will have some positive distance. Based on these observations, it's not possible for the average distance to be exactly 2, as there should be some value strictly between 0 and 2. Therefore, the statement is false.

Part c - Evaluating the integral in Cartesian and polar coordinates

We are given the integral $\int_{0}^{1} \int_{0}^{\sqrt{1-y^{2}}} e^{x^{2}+y^{2}} d x d y$ and asked to determine if it's easier to evaluate in polar coordinates than Cartesian coordinates. Let's first transform the integral into polar coordinates. Recall that \(x^2 + y^2 = r^2\) and that the region of integration is an area within the first quadrant of the unit circle. So, the corresponding polar representation is \(0 \le r \le 1\) and \(0 \le \theta \le \frac{\pi}{2}\). Note that in polar coordinates, the area differential element \(dA = r d\theta dr\). Transforming the integral, we get: $\int_{0}^{\pi/2} \int_{0}^{1} e^{r^2} \cdot r d r d \theta$. Now, comparing the Cartesian and polar integral forms, we can see that the expression inside the integral is simpler in the polar coordinates (\(e^{r^2}\) compared to \(e^{x^2+y^2}\)). However, integrating \(e^{r^2}\) with respect to \(r\) is still not an easy task in both coordinate systems, and none of them leads to an elementary function. Consequently, neither of the coordinate systems makes the integration straightforward. Therefore, the statement is false.

Key concepts

Double Integral

In calculus, a double integral refers to the integration of a function over a two-dimensional area. It's denoted by \( \iint \), and in Cartesian coordinates, for a function \( f(x, y) \), it can be written as \( \iint_R f(x, y) \,dA \) where \( R \) is the region of integration and \( dA = dx\,dy \) is the differential area element. This process accumulates all the infinitesimal contributions of \( f(x, y) \) over the area \( R \).

In the context of our exercise, the integral \( \iint_{R}\left(x^{2}+y^{2}\right) d A \) is an example of a double integral over the unit disk \( R \) centered at the origin. Key to evaluating double integrals is choosing the right coordinates for the given problem; it simplifies the integral both conceptually and computationally.

When using polar coordinates for circular or radial symmetric regions, we benefit from a natural fit between the symmetry of the region and the coordinates used. This can transform a challenging integral into a more straightforward calculation.

Cartesian Coordinates

The Cartesian coordinate system is a two-dimensional system in which each point is specified by a pair of numerical coordinates, which are the perpendicular distances to the point from two fixed perpendicular directed lines, measured in the same unit of length. These lines are called axes and labeled as the \( x \) and \( y \) axis.

The system is essential for much of algebra and calculus, especially for problems involving geometric figures like lines, curves, and regions in the plane. Cartesian coordinates simplify the expression of geometric shapes and can be generalized to higher dimensions.

However, there are certain kinds of problems, like those involving circular or spherical shapes where Cartesian coordinates may not be the most efficient way to proceed, leading to the use of other coordinate systems, such as polar coordinates, to facilitate the integration process.

Average Distance in Space

The average distance in space pertains to finding a mean value of distances from a set of points to a particular point, typically the origin. For a finite set of points in space, the average distance is calculated by summing up all distances and dividing by the number of points. However, for a continuous set of points, such as those on a line, area, or volume, the concept of an average distance is formulated via integral calculus.

In the example given in our textbook exercise, we're concerned with the average distance of points on a hemisphere to the origin. Calculating this distance involves integrating over the hemisphere surface and dividing by the hemisphere's area. Considering symmetry and spatial distribution helps infer whether the average distance aligns with certain intuitive bounds, though actual calculation can be complex.

Transformation of Coordinates

The transformation of coordinates is a process of converting a set of coordinates from one system to another. In calculus, this is commonly practiced when converting problems from Cartesian to polar coordinates, or vice versa.

During such transformations, not only do the coordinates change, but also the differential elements used for integration—like in the equation \( dA = r dr d\theta \) when converting the differential area element from Cartesian \( dx dy \) to polar coordinates. These transformations are indispensable when evaluating integrals over regions where one coordinate system provides a more simplified form of the function or bounds of integration than the other.

In the context of our example, for the double integral involving \( e^{x^{2}+y^{2}} \) over a circle quadrant, transforming to polar coordinates makes expressing the limits of integration much easier, but it does not necessarily simplify the actual integration, as the integral of \( e^{r^2} \) remains complex in any coordinates.