Question

Determine whether the following statements are true and give an explanation or counterexample. a. Let \(R\) be the unit disk centered at \((0,0) .\) Then \(\iint_{R}\left(x^{2}+y^{2}\right) d A=\int_{0}^{2 \pi} \int_{0}^{1} r^{2} d r d \theta\) b. The average distance between the points of the hemisphere \(z=\sqrt{4-x^{2}-y^{2}}\) and the origin is 2 (no integral needed). c. The integral \(\int_{0}^{1} \int_{0}^{\sqrt{1-y^{2}}} e^{x^{2}+y^{2}} d x d y\) is easier to evaluate in polar coordinates than in Cartesian coordinates.

Short answer

Question: Determine which of the following statements are true or false. a. The given double integral for the unit disk centered at (0, 0) is true when set up in polar coordinates. b. The average distance between points on the hemisphere z = sqrt(4 - x^2 - y^2) and the origin is 2 without necessity for an integral. c. The integral below would be easier to evaluate in polar coordinates: $$\int_{0}^{1} \int_{0}^{\sqrt{1-y^{2}}} e^{x^{2}+y^{2}} d x d y$$ Answer: a. True; b. False; c. False

Step-by-step solution

Set up the integral in polar coordinates

Let's set up the given double integral for the unit disk centered at \((0, 0)\) in polar coordinates. To convert the rectangular coordinates to polar coordinates, use the substitutions \(x = r\cos{\theta}\), \(y = r\sin{\theta}\), and \(dA = r dr d\theta\). Therefore, the given double integral becomes: $$\iint_{R}\left(x^{2}+y^{2}\right) d A=\int_{0}^{2 \pi} \int_{0}^{1} r^{2} d r d \theta$$

Verify the statement

Since we used the substitutions to convert rectangular coordinates to polar coordinates and set up the integral correctly, the statement in part (a) is true. ##Statement b##

Calculate the average distance

For the hemisphere \(z = \sqrt{4 - x^2 - y^2}\), we know that the radius is 2. The average distance between the points on the hemisphere and the origin is the same as half the radius. Hence, the average distance is 1. Since the average distance is not 2, the statement in part (b) is false. ##Statement c##

Analyze the given integral

The given integral is: $$\int_{0}^{1} \int_{0}^{\sqrt{1-y^{2}}} e^{x^{2}+y^{2}} d x d y$$

Assess the complexity

In this case, the integral is actually easier to evaluate in Cartesian coordinates because the integrand is an elementary function with no quadratic factors. Therefore, it is not necessary to change the coordinates to polar coordinates. In conclusion, the statement in part (c) is false.

Key concepts

Double Integrals

Double integrals, just like single integrals, are used to compute the accumulation of quantities over a region. However, instead of a line, double integrals evaluate over a two-dimensional area. This makes them particularly valuable in multivariable calculus.
Double integrals are expressed as \( \iint_{R} f(x, y) \, dA \), where \( R \) is the region over which you are integrating. The function \( f(x, y) \) in the integrand often represents physical quantities like density, and the differential area element \( dA \) can be expressed in different forms depending on the chosen coordinate system.
When computing double integrals in Cartesian coordinates, \( dA \) is expressed as \( dx \, dy \) or \( dy \, dx \), depending on the order of integration, over the defined area \( R \). For problems where the region is symmetric or circular, such as a disk, it is more convenient to perform integrations in polar coordinates.

Polar Coordinates

Polar coordinates offer an advantageous transformation in double integrals, particularly for circular or radial symmetry. Unlike Cartesian coordinates, which use \(x\) and \(y\) values, polar coordinates describe points based on their distance from the origin (radius \(r\)) and the angle \(\theta\) from the positive x-axis.
Using polar coordinates requires the transformations \( x = r \cos{\theta} \) and \( y = r \sin{\theta} \). The area element \( dA \) shifts from \( dx \, dy \) in Cartesian to \( r \, dr \, d\theta \), which accounts for the small change in area with increasing radius.
This makes evaluating integrals over circular regions particularly straightforward, as it simplifies setting the bounds and automatically includes the changes in area. When the problem involves circular or radial features, polar coordinates simplify the bounds, often transforming complex integrals into more tractable forms. This is evident in solving the integral of \( x^2 + y^2 \) over a unit disk, where it conveniently becomes \( r^2 \).

Coordinate Transformation

Coordinate transformation is a pivotal concept when it comes to simplifying the evaluation of double integrals. It involves changing from one coordinate system to another (such as from Cartesian to polar), which can make complex integrals easier to solve. This transformation is especially beneficial when the region of integration itself aligns naturally with one coordinate system over another.
To perform a coordinate transformation, you substitute the known relationships between the coordinates (e.g., \(x = r \cos{\theta}\), \(y = r \sin{\theta}\)) into the integral. It is crucial to adjust the area element, acknowledging the change in geometry using the Jacobian determinant. For polar transformations, the Jacobian simplifies to \( r \, dr \, d\theta \).
This approach is not only for simplifying computations but also for ensuring accuracy when evaluating integrals over complex shapes or when the function itself is influenced by the geometry of the coordinate system. For example, the transformation in the unit disk scenario efficiently allows the integration of \(r^2\) in polar coordinates, which is easier compared to direct evaluation in Cartesian coordinates over circular domains.