Problem 9 - Thomas Calculus

Question

In Exercises \(9-22,\) change the Cartesian integral into an equivalent polar integral. Then evaluate the polar integral. $$ \int_{-1}^{1} \int_{0}^{\sqrt{1-x^{2}}} d y d x $$

Step-by-Step Solution

Answer

The value of the integral is \( \frac{\pi}{2} \).

Step 1: Understand the Cartesian Integral Limits

The given Cartesian integral is \( \int_{-1}^{1} \int_{0}^{\sqrt{1-x^2}} dy \, dx \). This means \( x \) ranges from -1 to 1, while \( y \) ranges from 0 to \( \sqrt{1-x^2} \). These limits describe the upper half of a circle with radius 1 centered at the origin.

Step 2: Convert the Cartesian Integral to Polar Coordinates

In polar coordinates, the relationship between Cartesian coordinates \( (x, y) \) and polar coordinates \( (r, \theta) \) is given by \( x = r \cos\theta \) and \( y = r \sin\theta \). The region described by the Cartesian limits is a semicircle with radius 1 on the upper half. In polar coordinates, \( r \) ranges from 0 to 1, and \( \theta \) ranges from 0 to \pi \.

Step 3: Express the Differential Area in Polar Coordinates

In polar coordinates, the differential area element \( dy \, dx \) becomes \( r \, dr \, d\theta \). Substituting this into the integral, we get \( \int_{0}^{\pi} \int_{0}^{1} r \, dr \, d\theta \).

Step 4: Evaluate the Inner Integral with respect to \( r \)

The inner integral \( \int_{0}^{1} r \, dr \) needs to be evaluated. This integral evaluates to \[ \frac{r^2}{2} \Bigg|_{0}^{1} = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2}. \]

Step 5: Evaluate the Outer Integral with respect to \( \theta \)

Now that the inner integral evaluates to \( \frac{1}{2} \), the outer integral is \( \int_{0}^{\pi} \frac{1}{2} \, d\theta \). Evaluating this gives \[ \frac{1}{2} \theta \Bigg|_{0}^{\pi} = \frac{1}{2} \pi - \frac{1}{2} \times 0 = \frac{\pi}{2}. \]

Step 6: State the Final Result

The value of the polar integral is \( \frac{\pi}{2} \), which is the same as the value of the original Cartesian integral.

Key Concepts

Cartesian Coordinates

The concept of Cartesian coordinates is a cornerstone of mathematics, especially in the field of geometry. Cartesian coordinates describe a point in space with reference to a grid. This grid consists of two perpendicular lines, the x-axis (horizontal) and the y-axis (vertical).

• The point's position is determined by two numbers: \( (x, y) \).
• Here, \( x \) represents the point's distance from the y-axis, while \( y \) signifies the distance from the x-axis.

By specifying both \( x \) and \( y \), you can locate any point in the plane, making Cartesian coordinates very useful for graphing equations and describing geometric figures. In this exercise, the Cartesian coordinates describe the range of \( x \) from -1 to 1 and \( y \) from 0 to \( \sqrt{1-x^2} \). This forms a half-circle, which leads us to transform this into polar coordinates for easier integration.

Integral Calculus

Integral calculus is a branch of mathematics that deals with integrals and their applications. An integral can be thought of as the area under a curve or a more generalized concept of accumulation.

• Indefinite integrals provide a family of functions all described by the accumulation function.
• Definite integrals have bounds, where you evaluate the area under the curve between two points.

The integral we are dealing with in this exercise is definite and aims to find the area inside a semicircle. Understanding the concepts of limits and accumulation is key to solving such integrals. As shown, the Cartesian integral is transformed into a polar integral to ease the calculation involving semicircular regions.

Coordinate Transformation

Coordinate transformation is a process used to switch from one coordinate system to another. This is particularly useful in problems where a change of perspective simplifies the calculations.

• Cartesian to polar transformation uses the relations \(x = r \cos \theta\) and \(y = r \sin \theta\).
• Polar coordinates \( (r, \theta) \) describe points in terms of a radius and angle.

In the exercise, transforming the integral from Cartesian coordinates \( (x, y) \) to polar coordinates \( (r, \theta) \) allows us to simplify the evaluation, especially because the bounds describe a semicircle. This form makes the integral easier to handle because the region’s symmetry is better expressed in polar form.

Semicircle

A semicircle is a geometric shape that represents half of a circle. It is formed when a circle is cut by a diameter, resulting in two equal semicircles.

• The standard equation of a circle with radius 1 centered at the origin is \(x^2 + y^2 = 1\).
• In the upper half-plane, the semicircle is described by \(y = \sqrt{1-x^2}\).

In this exercise, the semicircle region is integral to understanding both coordinate transformation and integration limits. The semicircle's symmetry makes it easier to convert the Cartesian integral into polar form, leading to a simplified calculation with bounds \( r \) from 0 to 1 and \( \theta \) from 0 to \( \pi \). This elegant form makes integration straightforward, underscoring the power of using symmetry and appropriate coordinate transformations in solving calculus problems.