Problem 27 - Calculus Early Transcendentals

Question

Sketch the given region of integration \(R\) and evaluate the integral over \(R\) using polar coordinates. $$\begin{aligned} &\iint_{R} \frac{d A}{\sqrt{16-x^{2}-y^{2}}}\\\ &R=\left\\{(x, y): x^{2}+y^{2} \leq 4, x \geq 0, y \geq 0\right\\} \end{aligned}$$

Step-by-Step Solution

Answer

Question: Calculate the double integral of \(\frac{1}{\sqrt{16-x^2-y^2}}\) over the region R, which is defined by the inequalities \(x^2 + y^2 \leq 4\), \(x \geq 0\), and \(y \geq 0\). Answer: The double integral of \(\frac{1}{\sqrt{16-x^2-y^2}}\) over the region R equals \(\pi\).

Step 1: Identify the Boundaries

We are given the inequalities \(x^2 + y^2 \leq 4\), \(x \geq 0\), and \(y \geq 0\). The inequality \(x^2 + y^2 \leq 4\) describes a circle centered at the origin with a radius of 2, with points both inside and on the circle included. The inequalities \(x \geq 0\) and \(y \geq 0\) restrict R to the first quadrant.

Step 2: Sketch the Region R

In the first quadrant, draw a circle with a radius 2 centered at the origin. The region R is the quarter of the circle located in the first quadrant.

Step 3: Change Cartesian Coordinates to Polar Coordinates

Now, we switch from Cartesian coordinates \((x, y)\) to polar coordinates \((r, \theta)\) by using the following transformations: \(x = r\cos{\theta}\), \(y = r\sin{\theta}\), and \(dA = r dr d\theta\). Then, the given integral becomes: $$\iint_{R} \frac{r dr d\theta}{\sqrt{16-r^{2}}}$$. From the inequalities \(x^2+ y^2 \leq 4\), \(x\geq 0\), and \( y\geq 0\), we can deduce that \(0 \leq r \leq 2\) and \(0 \leq \theta \leq \frac{\pi}{2}\)

Step 4: Evaluate the Integral

Now, we will evaluate the double integral in polar coordinates over the region R: $$\begin{aligned} \int_{0}^{\frac{\pi}{2}} \int_{0}^{2} \frac{r dr d\theta}{\sqrt{16-r^{2}}} &=\int_{0}^{\frac{\pi}{2}} d\theta \int_{0}^{2} \frac{r dr}{\sqrt{16-r^{2}}} \\ &=\int_{0}^{\frac{\pi}{2}} d\theta \left[-\sqrt{16-r^2}\right]_0^2 \\ &=\int_{0}^{\frac{\pi}{2}} d\theta \left(-\sqrt{16-2^2}+\sqrt{16-0}\right) \\ &=\int_{0}^{\frac{\pi}{2}} d\theta (2) \\ &=\left[ 2\theta\right]_0^{\frac{\pi}{2}} \\ &=\pi\,. \end{aligned}$$ Thus, the integral over the region R equals \(\pi\).

Key Concepts

Double Integral

A double integral is a form of multiple integral that allows one to integrate over a two-dimensional area. In mathematics, it's often used to find the volume beneath a surface or the area of a region in a plane. Essentially, when we use a double integral, we're adding up all the values of a function at points on a certain region or surface. Imagine laying a grid over a flat park and trying to figure out exactly how high each blade of grass is; a double integral would help you find the total height of the grass over the entire park.

When the function being integrated is as simple as one, as in the given exercise, the double integral calculates the area of the specified region. The first integral sums the values along an infinitesimal strip, and the second integral then adds up these strips to cover the entire region.

Cartesian to Polar Coordinate Transformation

The Cartesian to polar coordinate transformation is a way to switch from the Cartesian coordinate system (using x and y) to the polar coordinate system (using r and \(\theta\)). In polar coordinates, points are determined by how far away they are from the origin (that's the 'r' or radius) and the angle they make with the positive x-axis (that's the '\((\theta\)' or theta).

To convert, you use the equations \(x = r\cos(\theta)\) and \(y = r\sin(\theta)\). This conversion simplifies the integration process when dealing with circular or radial symmetry, as is the case for circles, spheres, and similar shapes. By transforming our integral to polar coordinates in the exercise, it becomes easier to evaluate, especially since the region of integration is a quarter of a circle.

Evaluating Integrals

Evaluating integrals is the process of calculating the numerical value of an integral, which might represent the area under a curve, the volume under a surface, or other physical quantities. The fundamental theorem of calculus is the key player here, providing the link between differentiation and integration. When evaluating definite integrals, we're usually finding the accumulation of quantities, like distance travelled given a velocity function, or mass, given a density function. In the context of polar coordinates, we have an additional step before the evaluation: converting the original function into terms of r and \(\theta\), and remembering to include \(r\) in the differential because area elements in polar coordinates are not rectangles, but rather pie-slice shaped.

Region of Integration Sketching

Sketching the region of integration is a visual step that aids in understanding the area over which the double integral is to be evaluated. This is particularly important when converting from Cartesian to polar coordinates, as it helps identify the bounds for \(r\) and \(\theta\).

In this exercise, the region is the part of the circle with radius 2 that lies in the first quadrant. By sketching this region, we are able to see that \(r\) ranges from 0 to 2 (from the center out to the edge of the circle) and \(\theta\) ranges from 0 to \(\frac{\pi}{2}\) (from the positive x-axis to the positive y-axis). Thus, sketching is not merely a preliminary step; it is integral to setting up the correct limits of integration for polar coordinates.