Problem 23 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate by using polar coordinates. Sketch the region of integration first. $$ \int_{0}^{1} \int_{0}^{\sqrt{1-x^{2}}}\left(4-x^{2}-y^{2}\right)^{-1 / 2} d y d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{\pi}{4}(\sqrt{3} - 1) \).

Step 1: Understanding the Region of Integration

The inner integral is with respect to the variable \( y \) and varies from 0 to \( \sqrt{1-x^2} \). This covers the upper semi-circle from the origin since \( y = \sqrt{1-x^2} \) describes a semi-circle of radius 1 centered at the origin in the xy-plane. The outer integral varies \( x \) from 0 to 1.

Step 2: Sketch the Region of Integration

The region of integration is a quarter-circle in the first quadrant of the xy-plane, since \( y \) ranges from 0 to \( \sqrt{1-x^2} \) and \( x \) ranges from 0 to 1, creating a quarter section of a circle with radius 1.

Step 3: Switch to Polar Coordinates

Convert the rectangular coordinates to polar coordinates using the transformations \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \). The region described originally by \( 0 \leq x \leq 1 \) and \( 0 \leq y \leq \sqrt{1-x^2} \) is bounded in polar coordinates by \( 0 \leq r \leq 1 \) and \( 0 \leq \theta \leq \frac{\pi}{2} \).

Step 4: Re-evaluate the Integral in Polar Coordinates

Substitute \( x = r \cos(\theta) \), \( y = r \sin(\theta) \), and \( dx \, dy = r \, dr \, d\theta \) into the integral. The expression becomes: \[\int_{0}^{\pi/2} \int_{0}^{1} (4-r^2)^{-1/2} \cdot r \, dr \, d\theta\]

Step 5: Solve the Inner Integral with Respect to r

Recognize that the integral \( \int_{0}^{1} r(4-r^2)^{-1/2} \, dr \) can be solved by substitution. Let \( u = 4 - r^2 \), then \( du = -2r \, dr \). Rewrite the limits for \( u \) as 4 to 3, based on the substitution. The integral simplifies and evaluates as: \[ \frac{1}{2} \int_{4}^{3} u^{-1/2} \cdot (-1/2) du = \frac{1}{4} [2\sqrt{u}]_{4}^{3} = \frac{1}{4} (2\sqrt{3} - 2) = \frac{1}{2}(\sqrt{3} - 1)\]

Step 6: Solve the Outer Integral with Respect to θ

Finally, compute the outer integral with respect to \( \theta \): \[\int_{0}^{\pi/2} \frac{1}{2}(\sqrt{3} - 1) \, d\theta = \frac{1}{2}(\sqrt{3} - 1) \cdot \frac{\pi}{2} = \frac{\pi}{4}(\sqrt{3} - 1)\]

Step 7: Final Answer

The evaluated integral in polar coordinates is \( \frac{\pi}{4}(\sqrt{3} - 1) \).

Key Concepts

Quarter-Circle Region

Understanding the concept of a quarter-circle region is crucial when shifting from rectangular to polar coordinates for integration. A quarter-circle is essentially a 90-degree section of a full circle. In the first quadrant of the xy-plane, this portion appears when both x and y are positive.
The original exercise involves the integral: \[\int_{0}^{1} \int_{0}^{\sqrt{1-x^{2}}}\left(4-x^{2}-y^{2}\right)^{-1 / 2} d y d x\]The bounds imply that x runs from 0 to 1, and y varies from 0 to the semi-circle with equation \( y = \sqrt{1-x^2} \). This describes a quarter-circle region with a radius of 1. Sketching this, you’ll see a rounded segment in the first quadrant which accurately defines the region over which the integration occurs. Recognizing this can assist in visualizing and shifting to polar coordinates later.

Double Integration

Double integration is the process of evaluating an integral iteratively, first over one variable and then the other. This technique is often employed for calculating areas, volumes, or other quantities that spread over an entire region.
In the original exercise, the integration process involved two variables: x and y. Initially, you focus on integrating with respect to y, keeping x fixed. The inner integral operation: \[\int_{0}^{\sqrt{1-x^{2}}}\left(4-x^{2}-y^{2}\right)^{-1 / 2} d y\]solves part of the problem by reducing it to a single integral over x. After evaluating the inner integral, the outer integral is performed over x, to complete the evaluation. By structuring the problem this way, the integration over a two-dimensional space becomes manageable and logical.

Substitution in Integration

Substitution is a powerful method to simplify integrals by changing variables. This technique is particularly helpful when encountering complex expressions that are otherwise tough to handle with straightforward integration.
During the calculation, after transforming to polar coordinates, the integral of the form: \[\int_{0}^{1} r(4-r^2)^{-1/2} \, dr\]is simplified using substitution. By letting \( u = 4 - r^2 \), you derive \( du = -2r \, dr \). This transforms the original variable r into a simpler expression in terms of u, making the integration more tractable: \[\frac{1}{2} \int_{4}^{3} u^{-1/2} \cdot (-1/2) \, du\]. Substitution not only simplifies the calculus but sometimes reveals more straightforward numeric results, leading to accurate solutions.

Trigonometric Transformations

Trigonometric transformations involve converting Cartesian coordinates (x, y) into polar coordinates (r, \(\theta\)). This is particularly useful in problems with circular symmetry, where the boundaries align naturally with polar coordinates.
In the problem, the transition is made using the relations:

• \( x = r \cos(\theta) \)
• \( y = r \sin(\theta) \)

By converting the differentials, the integral transforms into: \[\int_{0}^{\pi/2} \int_{0}^{1} (4-r^2)^{-1/2} \cdot r \, dr \, d\theta\]which manifests in polar form. Trigonometric transformations adjust the view from rectangular regions to circular, easing the mathematical process. It especially simplifies integration limits, modifying what might appear as complicated patterns to more logical numerical bounds corresponding with angles and radii.