Problem 44 - Calculus Early Transcendentals

Question

Evaluate the following integrals in spherical coordinates. $$\int_{0}^{2 \pi} \int_{0}^{\pi / 4} \int_{1}^{2 \sec \varphi}\left(\rho^{-3}\right) \rho^{2} \sin \varphi d \rho d \varphi d \theta$$

Step-by-Step Solution

Answer

Question: Evaluate the given integral in spherical coordinates: $$\int_{0}^{2 \pi} \int_{0}^{\pi / 4} \int_{1}^{2 \sec \varphi}\left(\rho^{-3}\right) \rho^{2} \sin \varphi d \rho d \varphi d \theta$$ Answer: The integral evaluates to $$(F(\frac{\pi}{4}) - F(0))(2 \pi)$$ where $$F(\phi) = \int -\ln (2 \sec \varphi) \sin \varphi d \varphi$$ and F(0) and \(F(\frac{\pi}{4})\) are constants.

Step 1: Integrate with respect to ρ

First, we will integrate the expression with respect to ρ. The integral is: $$\int_{1}^{2 \sec \varphi}\left(\rho^{-3}\right) \rho^{2} \sin \varphi d \rho$$ We can simplify the integrand before proceeding: $$\rho^{-3}\cdot \rho^2 = \rho^{-1}$$ Now, we can integrate: $$\int_{1}^{2 \sec \varphi} \rho^{-1} d \rho$$ Using the power rule for integration, we have: $$[- \ln |\rho|]_{1}^{2 \sec \varphi}$$ This evaluates to: $$-\ln |2 \sec \varphi| + \ln |1|$$ Since the natural logarithm of 1 is 0, the result simplifies to: $$-\ln (2 \sec \varphi)$$ Now, we have the innermost integral evaluated.

Step 2: Integrate with respect to φ

Next, we will integrate the expression with respect to φ. The integral is: $$\int_{0}^{\pi / 4} -\ln (2 \sec \varphi) \sin \varphi d \varphi$$ There is no elementary antiderivative for the function within the integral, so we will represent it with a general expression with respect to φ: $$F(\phi) = \int -\ln (2 \sec \varphi) \sin \varphi d \varphi$$ We have nothing further to simplify or evaluate, so our integral becomes: $$[F(\phi)]_{0}^{\pi / 4}$$ This evaluates to: $$F(\frac{\pi}{4}) - F(0)$$

Step 3: Integrate with respect to θ

Finally, we will integrate the expression with respect to θ. The integral is: $$\int_{0}^{2 \pi} F(\frac{\pi}{4}) - F(0) d \theta$$ Since \(F(\frac{\pi}{4})\) and \(F(0)\) are constants with respect to θ, we can simply evaluate the integral as: $$(F(\frac{\pi}{4}) - F(0))\int_{0}^{2 \pi} d \theta$$ Which evaluates to: $$(F(\frac{\pi}{4}) - F(0))( \theta|_{0}^{2\pi})$$ This evaluates to: $$(F(\frac{\pi}{4}) - F(0))(2 \pi)$$ Thus, the integral we want to evaluate in spherical coordinates is: $$\int_{0}^{2 \pi} \int_{0}^{\pi / 4} \int_{1}^{2 \sec \varphi}\left(\rho^{-3}\right) \rho^{2} \sin \varphi d \rho d \varphi d \theta = (F(\frac{\pi}{4}) - F(0))(2 \pi)$$

Key Concepts

Multiple Integration

Multiple integration represents the extension of single-variable integral calculus to functions of several variables. When dealing with functions that depend on more than one variable, the process of integration can be carried out successively over each variable one at a time. In the context of the given problem, the integral represents a volume in a three-dimensional space using spherical coordinates. Here, the order of integration is crucial because the limits for each subsequent integral can depend on the variables of integration that come before it.

For instance, the integral given in the exercise first considers the radial parameter \( \rho \), followed by the angle \( \varphi \), and finally the rotational angle \( \theta \). By integrating in a specific sequence—\( \rho \) first, followed by \( \varphi \) and \( \theta \)—we are essentially summing up the contributions of small volume elements in spherical coordinates to find a total volume or mass.

Power Rule for Integration

The power rule for integration is a fundamental technique used to evaluate integrals of power functions. It states that the integral of \( x^n \) with respect to x for any real number n (except -1) is given by \( \frac{x^{n+1}}{n+1} \) plus a constant of integration. Importantly, when \( n = -1 \), the integral of \( x^{-1} \) is the natural logarithm function \( \ln|x| \)—which is precisely what we see in the solution of the innermost integral \( \rho^{-1} \).

In our example, after simplifying the integrand, the integral becomes \( \int \rho^{-1} d\rho \) leading to \( -\ln |\rho| + C \), where C is the constant of integration. This application of the power rule is the first step in solving the problem. It is an essential skill for students to master and is widely used in calculus for various applications.

Natural Logarithm Properties

The natural logarithm, denoted as \( \ln \) and based on Euler's number \( e \), has properties that make it particularly useful in calculus, including integration involving exponentials. One key property is that the natural logarithm of 1 is 0, as mathematically expressed as \( \ln(1) = 0 \). This property streamlines calculations, especially in definite integrals where one limit of integration can be 1.

Another important property is the logarithm of a product, which states that \( \ln(ab) = \ln(a) + \ln(b) \). However, in the given exercise reduction of the logarithmic expression comes from the fact that one of the terms, \( \ln(1) \)—from the evaluated bounds of the integral—simplifies to 0, hence leaving us with the expression for the intermediate integral in the form of \( -\ln(2 \sec(\varphi)) \). Understanding these properties is essential when solving integrals that result in logarithmic functions, as it aids in simplifying the final expression.