Problem 3 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate the iterated integrals. \(\int_{0}^{\pi} \int_{0}^{\sin \theta} r^{2} d r d \theta\)

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{1}{2} \).

Step 1: Understand the Problem

We are given an iterated integral \( \int_{0}^{\pi} \int_{0}^{\sin \theta} r^{2} \, dr \, d\theta \). The integral is evaluated with respect to \( r \) first and then with respect to \( \theta \). The limits for \( r \) are 0 to \( \sin \theta \), and the limits for \( \theta \) are 0 to \( \pi \).

Step 2: Integrate with respect to r

First, we integrate \( r^2 \) with respect to \( r \). The integral is \( \int r^2 \, dr = \frac{r^3}{3} + C \). We apply the limits from 0 to \( \sin \theta \).

Step 3: Apply Limits for r

Substitute the limits into the result of integration with respect to \( r \):\[ \left. \frac{r^3}{3} \right|_{0}^{\sin \theta} = \frac{(\sin \theta)^3}{3} - \frac{0^3}{3} = \frac{(\sin \theta)^3}{3} \].

Step 4: Reduce the Integral with respect to θ

Now that we have evaluated the inner integral, the integral becomes \( \int_{0}^{\pi} \frac{(\sin \theta)^3}{3} \, d\theta \). We can factor out the \( \frac{1}{3} \) since it's a constant:\[ \frac{1}{3} \int_{0}^{\pi} (\sin \theta)^3 \, d\theta \].

Step 5: Evaluate Integral of (sin θ)^3

To integrate \((\sin \theta)^3\), we use the identity \((\sin^n \theta)\) for reducing, or use integration by parts. For this, use the identity for power reduction: \(\sin^3 \theta = \frac{3\sin \theta - \sin 3\theta}{4}\), thus:\[ \int (\sin \theta)^3 \, d\theta = \int \frac{3\sin \theta - \sin 3\theta}{4} \, d\theta \].

Step 6: Simplify and Integrate Reduced Expression

We proceed by integrating both terms separately:- \( \int \frac{3\sin \theta}{4} d\theta = -\frac{3\cos \theta}{4} + C_1 \).- \( \int \frac{-\sin 3\theta}{4} d\theta = \frac{\cos 3\theta}{12} + C_2 \).Therefore,\[ \int_{0}^{\pi} (\sin \theta)^3 \, d\theta = \left[ -\frac{3\cos \theta}{4} + \frac{\cos 3\theta}{12} \right]_{0}^{\pi} \].

Step 7: Apply Limits for θ

Substituting the limits, we have:\[ \left[ -\frac{3\cos \theta}{4} + \frac{\cos 3\theta}{12} \right]_{0}^{\pi} = \left( -\frac{3\cos \pi}{4} + \frac{\cos 3\pi}{12} \right) - \left( -\frac{3\cos 0}{4} + \frac{\cos 0}{12} \right) \].Calculating:- At \( \theta = \pi \), it evaluates to \( -\frac{3(-1)}{4} - \frac{(-1)}{12} = \frac{3}{4} + \frac{1}{12} = \frac{9}{12} + \frac{1}{12} = \frac{10}{12} = \frac{5}{6} \).- At \( \theta = 0 \), it evaluates to \( -\frac{3(1)}{4} + \frac{1}{12} = -\frac{3}{4} + \frac{1}{12} = -\frac{9}{12} + \frac{1}{12} = -\frac{8}{12} = -\frac{2}{3} \).Thus, the definite integral is \( \frac{5}{6} - \left(-\frac{2}{3}\right) = \frac{5}{6} + \frac{2}{3} = \frac{9}{6} = \frac{3}{2} \).

Step 8: Finalize Calculation

The integral becomes:\[ \frac{1}{3} \times \frac{3}{2} = \frac{3}{6} = \frac{1}{2} \].Thus, the iterated integral evaluates to \( \frac{1}{2} \).

Key Concepts

Double Integrals

Double integrals enable us to calculate the volume under a surface in a three-dimensional space. They extend the concept of a single integral, from integrating along a line, to integrating over a two-dimensional area.

In a double integral, we set up an iterated process where we first integrate with respect to one variable, while treating the other variable as a constant. After completing this, we integrate again, this time treating the first variable as the constant.

When working with double integrals:

• We often can think of the first integration as finding the area under a curve for a fixed point of the other variable.
• The limits of integration for each variable define the boundary of the region over which we are integrating.
• The process is helpful in fields such as physics and engineering for calculating mass, center of mass, or electric/current flow over a region.

In the given problem, we evaluated \(\int_{0}^{\pi} \int_{0}^{\sin \theta} r^2 dr d\theta\), examining how the value of the integral evolves as we move in the \(\theta\) direction while accumulating the value of \(r^2\) along the way.

Integration by Parts

Integration by parts is a powerful technique used to solve integrals of the product of two functions. Derived from the product rule of differentiation, it helps us turn an integral into a more manageable form.

The formula for integration by parts is:

• \(\int u \, dv = uv - \int v \, du\)

Here, \(u\) and \(dv\) are parts of the original integral. By differentiating \(u\) to get \(du\) and integrating \(dv\) to get \(v\), we can reformulate a complex problem into subtraction of a simpler integral.

In our exercise, we indirectly applied the concept when integrating trigonometric functions by manipulating power of sine function through identities, letting us simplify our integral for easier computation.

Trigonometric Integrals

Trigonometric integrals involve integrating powers or products of sine and cosine functions. They often require clever manipulations like using trigonometric identities or substitutions.

For example, when integrating \((\sin \theta)^3\), as in our exercise, we employed a power-reduction identity:

• \(\sin^3 \theta = \frac{3\sin \theta - \sin 3\theta}{4}\)

This step simplifies the integral into terms of lower powers, enabling us to integrate each component separately.

Such techniques allow breaking down the integral into sums of fractions of simpler, basic trigonometric integrals, easing the computation of otherwise difficult scenarios.

Definite Integrals

Definite integrals help us evaluate the net area under a curve between two limits. They provide the means to compute areas, volumes, displacement, and other quantities over a specific interval.

The process involves:

• Integrating the function with respect to its variable.
• Applying the fundamental theorem of calculus to evaluate at the upper and lower limits.
• Subtracting these results to find the final integrated value.

In our problem, we dealt with \(\theta\) from 0 to \(\pi\). After simplifying \(\sin^3 \theta\) using trigonometric identities, we integrated each term separately over this interval.

Applying definite limits involves substituting the upper and lower limits into the antiderivative, as shown when substituting \(\theta\) values, leading us to the final result of \(\frac{1}{2}\). Overall, definite integrals give a precise numerical value that quantifies accumulated quantities over defined bounds.