Problem 30 - Thomas Calculus

Question

Gives an integral over a region in a Cartesian coordinate plane. Sketch the region and evaluate the integral. \begin{equation} \int_{0}^{1} \int_{0}^{\sqrt{1-s^{2}}} 8 t d t d s \quad(\text { the st-plane }) \end{equation}

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{8}{3} \).

Step 1: Understand the Region

The integral is given in the form \( \int_{0}^{1} \int_{0}^{\sqrt{1-s^{2}}} 8 t \; dt \; ds \). This represents a region in the \( st \)-plane. The outer integral \( s \) ranges from 0 to 1 and for each fixed \( s \), \( t \) ranges from 0 to \( \sqrt{1-s^{2}} \). Thus, the region is bounded by \( t = 0 \), the curve \( t = \sqrt{1-s^{2}} \) and \( s = 0 \) to 1.

Step 2: Sketch the Region

The curve \( t = \sqrt{1-s^{2}} \) represents the upper half of the unit circle centered at the origin with radius 1. Thus, in the \( st \)-plane, the region is a quarter of a circle in the first quadrant, bounded by \( t = 0 \), \( s = 0 \), and the curve \( t = \sqrt{1-s^{2}} \).

Step 3: Evaluate the Inner Integral

Evaluate the integral with respect to \( t \): \[ \int_{0}^{\sqrt{1-s^{2}}} 8t \; dt \]Using the power rule, find the antiderivative: \[ = \left[ 4t^2 \right]_{0}^{\sqrt{1-s^{2}}} \]= \[ 4 (\sqrt{1-s^{2}})^2 - 0 \] = \[ 4(1-s^{2}) \]

Step 4: Evaluate the Outer Integral

Integrate the result from Step 3 with respect to \( s \):\[ \int_{0}^{1} 4(1-s^{2}) \; ds \]Separate the integral:\[ = \int_{0}^{1} 4 \; ds - \int_{0}^{1} 4s^{2} \; ds \]Calculate the first integral:\[ = [ 4s ]_{0}^{1} = 4 \]Calculate the second integral using the power rule:\[ = \left[ \frac{4s^3}{3} \right]_{0}^{1} = \frac{4}{3} \]Subtract the second result from the first:\[ 4 - \frac{4}{3} = \frac{12}{3} - \frac{4}{3} = \frac{8}{3} \]

Step 5: Conclude the Calculation

After computing both the inner and outer integrals, we find that the value of the double integral over the region is \( \frac{8}{3} \).

Key Concepts

Cartesian Coordinate Plane

The Cartesian coordinate plane is an essential concept in understanding double integrals. It is a two-dimensional plane defined by a horizontal axis, usually labeled as the x-axis, and a vertical axis, known as the y-axis. In the case of the given problem, the axes are labeled as s and t, making it the st-plane. This plane allows us to graphically represent and analyze mathematical functions and their integrals visually.

• Coordinates are given in pairs (s, t), indicating positions on the plane.
• Each point on the plane corresponds to an intersection of the s and t values.

When working with a double integral, the Cartesian coordinate plane helps to specify the region of integration. By visualizing these regions, students can better grasp the limits of integration and the overall setup of the problem. To effectively use this plane, one must understand how curves and lines divide and define regions, such as a circle or half-circle, which is a frequent occurrence in exercises involving trigonometric functions.

Region of Integration

The concept of the region of integration is crucial when evaluating double integrals. It refers to the specific area over which integration occurs in the Cartesian coordinate plane.
In the exercise provided, the region of integration is defined by the limits of the integrals:

• For each fixed value of s, t ranges from 0 to \( \sqrt{1-s^2} \).
• The external limit for s goes from 0 to 1.

This setup implies that the region of integration is confined to a part of the plane where these conditions are met. For the problem, this results in a quarter circle bounded in the first quadrant by t = 0, s = 0, and the curve \( t = \sqrt{1-s^2} \). Understanding this region helps when sketching the area to provide a visual interpretation of what the integral represents.

Unit Circle

The unit circle is a fundamental concept in trigonometry and calculus. It refers to a circle with a radius of one, centered at the origin of the Cartesian coordinate plane. In the context of double integrals, sections of the unit circle often appear as part of the region of integration.
In the given exercise, the curve \( t = \sqrt{1-s^2} \) derives from the unit circle equation \( s^2 + t^2 = 1 \). This means:

• It is the upper half of the unit circle since t takes on non-negative values.
• It extends from s = 0 to s = 1, forming a quarter circle in the first quadrant.

Recognizing the unit circle in a problem helps in visualizing and sketching the region of integration. Moreover, it connects to deeper trigonometric concepts, as sin and cos functions are heavily based on unit circle geometry.

Antiderivative

Antidifferentiation, or finding an antiderivative, is the reverse process of differentiation. It is crucial when dealing with integrals, as it allows you to determine the area under a curve or evaluate definite integrals. In the context of double integrals, finding the antiderivative is part of calculating the integral over a given region.
For the inner integral in the exercise \[ \int_{0}^{\sqrt{1-s^{2}}} 8t \; dt \] We first calculate the antiderivative of 8t: \[ \left[ 4t^2 \right] \] Then, evaluate the bounds from 0 to \( \sqrt{1-s^{2}} \): \[ 4(\sqrt{1-s^{2}})^2 - 0 = 4(1-s^{2}) \] Finding an antiderivative is an essential skill in calculus, enabling students to move from simply understanding rates of change to applying that understanding in evaluating integrals and understanding areas and accumulations.