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Chapter 14: Problem 2

Evaluate the iterated integral. $$ \int_{0}^{\pi} \int_{0}^{1+\cos \theta} r d r d \theta $$

Short Answer

Expert verified
The iterated integral evaluates to \( \frac{3\pi}{4} \).

Step by step solution

01

Setup the Problem

The given integral is a double integral, which means we need to evaluate the integral with respect to one variable first, and then the other. The inner integral is \( \int_{0}^{1+\cos \theta} r \, dr \), and the outer integral is \( \int_{0}^{\pi} (\cdot) \, d\theta \).
02

Evaluate the Inner Integral

To evaluate the inner integral \( \int_{0}^{1+\cos \theta} r \, dr \), use the power rule for integration. Integrate \( r \) to get \( \frac{r^2}{2} \), then evaluate it from 0 to \( 1 + \cos \theta \). This gives: \[\int_{0}^{1+\cos \theta} r \, dr = \left[ \frac{r^2}{2} \right]_{0}^{1+\cos \theta} = \frac{(1+\cos \theta)^2}{2}.\]
03

Simplify the Result of the Inner Integral

Expand \((1+\cos \theta)^2\) to make the integration easier: \[(1+\cos \theta)^2 = 1 + 2\cos \theta + \cos^2 \theta.\]Therefore, \[\frac{(1+\cos \theta)^2}{2} = \frac{1 + 2\cos \theta + \cos^2 \theta}{2}.\]
04

Evaluate the Outer Integral

Now, integrate the expression \( \frac{1}{2} + \cos \theta + \frac{\cos^2 \theta}{2} \) with respect to \( \theta \) from 0 to \( \pi \):\[\int_{0}^{\pi} \left( \frac{1}{2} + \cos \theta + \frac{\cos^2 \theta}{2} \right) \, d\theta = \int_{0}^{\pi} \frac{1}{2} \, d\theta + \int_{0}^{\pi} \cos \theta \, d\theta + \frac{1}{2} \int_{0}^{\pi} \cos^2 \theta \, d\theta.\]
05

Solve Each Part Separately

1. Solve \( \int_{0}^{\pi} \frac{1}{2} \, d\theta = \left[ \frac{\theta}{2} \right]_{0}^{\pi} = \frac{\pi}{2}. \)2. Solve \( \int_{0}^{\pi} \cos \theta \, d\theta = [\sin \theta ]_{0}^{\pi} = 0. \)3. Use the identity \( \cos^2 \theta = \frac{1 + \cos 2\theta}{2} \) to transform the integral: \[ \int_{0}^{\pi} \cos^2 \theta \, d\theta = \int_{0}^{\pi} \frac{1 + \cos 2\theta}{2} \, d\theta = \left[ \frac{\theta}{2} + \frac{\sin 2\theta}{4} \right]_{0}^{\pi} = \frac{\pi}{2}. \] Hence, \( \frac{1}{2} \int_{0}^{\pi} \cos^2 \theta \, d\theta = \frac{\pi}{4}. \)
06

Combine the Results

Combine the solutions from each part to find the value of the outer integral:\[\frac{\pi}{2} + 0 + \frac{\pi}{4} = \frac{3\pi}{4}.\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Double Integral
In mathematics, a double integral allows us to compute integrals over a two-dimensional plane, which is particularly useful when working with areas, surfaces, and volumes. Essentially, you are performing integration twice: once with respect to one variable (like "r" in polar coordinates) and once with respect to another (such as "\(\theta\)").

For this exercise, the given iterated integral is \[ \int_{0}^{\pi} \int_{0}^{1+\cos \theta} r \, d r \, d \theta. \]Here's how it works:
  • Inner Integral: You start by integrating the function with respect to "r," treating "\(\theta\)" as a constant. After solving, you insert the limits for "r" to compute the result of the inner integral.

  • Outer Integral: Once the inner integral is solved, you move on to the outer integral with respect to "\(\theta\)," using the result from the inner integral as the function to be integrated.
This particular double integral is setup conveniently as polar coordinates, but we'll go into more detail in the next section.
Polar Coordinates
Polar coordinates are often used when dealing with problems involving circular or cylindrical symmetry. Instead of representing points using Cartesian coordinates \((x, y)\), polar coordinates use the radius "r" and angle "\(\theta\)."

The given integral \( \int_{0}^{\pi} \int_{0}^{1+\cos \theta} r \, d r \, d \theta \) suggests using a circular region. The limits for "r" and "\(\theta\)" correspond to a certain region in the polar plane.
  • Range for "\(\theta\)": From 0 to \(\pi\), which means you are sweeping out angles to fill half of the circle (upper semicircle).

  • Range for "r": From 0 to \(1 + \cos \theta\), which describes a limacon shape, a "heart" shape often found in polar coordinates.
The combination of these ranges describes the swirling region over which you're integrating.
Trigonometric Identity
Trigonometric identities are formulas that express relationships between the trigonometric functions, making complex problems more manageable.

In the solution, one key trigonometric identity used is \( \cos^2 \theta = \frac{1 + \cos 2\theta}{2} \). This identity helps transform the integral \( \int_{0}^{\pi} \cos^2 \theta \, d\theta \), making it easier to solve.
  • The transformation changes the power of "cosine" into simpler forms, involving terms you can integrate without additional identities.

  • Utilizing identities cleverly can reduce difficult integrals into simpler ones, just as it did by breaking down \( \cos^2 \theta \) into more manageable parts.
Applying such identities is crucial in solving trigonometric integrals, especially in iterated integrals involving polar coordinates.

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Most popular questions from this chapter

Evaluate the iterated integral. \(\int_{0}^{\pi / 4} \int_{0}^{1} \int_{0}^{x^{2}} x \cos y d z d x d y\)

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The tendency of a lamina to resist a change in rotational motion about an axis is measured by its moment of inertia about that axis. If a lamina occupies a region \(R\) of the \(x y\) -plane, and if its density function \(\delta(x, y)\) is continuous on \(R\), then the moments of inertia about the \(x\) -axis, the \(y\) -axis, and the \(z\) -axis are denoted by \(I_{x}, I_{y}\), and \(I_{z}\), respectively, and are defined by $$\begin{aligned}&I_{x}=\iint_{R} y^{2} \delta(x, y) d A, \quad I_{y}=\iint_{R} x^{2} \delta(x, y) d A \\\&I_{z}=\iint_{R}\left(x^{2}+y^{2}\right) \delta(x, y) d A \end{aligned}$$ Use these definitions. Consider the circular lamina that occupies the region described by the inequalities \(0 \leq x^{2}+y^{2} \leq a^{2}\). Assuming that the lamina has constant density \(\delta\), show that $$I_{x}=I_{y}=\frac{\delta \pi a^{4}}{4}, \quad I_{z}=\frac{\delta \pi a^{4}}{2}$$

Let \(G\) be the rectangular box defined by the inequalities $$\begin{aligned}&a \leq x \leq b, c \leq y \leq d, k \leq z \leq l . \text { Show that } \\ &\iiint_{G} f(x) g(y) h(z) d V \\ &\quad=\left[\int_{a}^{b} f(x) d x\right]\left[\int_{c}^{d} g(y) d y\right]\left[\int_{k}^{l} h(z) d z\right]\end{aligned}$$
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