Question

Changing order of integration If possible, write iterated integrals in spherical coordinates for the following regions in the specified orders. Sketch the region of integration. Assume that \(f\) is continuous on the region. $$\begin{aligned}&\int_{0}^{2 \pi} \int_{0}^{\pi / 4} \int_{0}^{4 \sec \varphi} f(\rho, \varphi, \theta) \rho^{2} \sin \varphi d \rho d \varphi d \theta \text { in the orders }\\\&d \rho d \theta d \varphi \text { and } d \theta d \rho d \varphi\end{aligned}$$

Short answer

$$\int_{0}^{2 \pi} \int_{0}^{\pi / 4} \int_{0}^{4 \sec\varphi} f(\rho, \varphi, \theta) \rho^{2} \sin \varphi d \rho d \varphi d \theta$$ The new integration bounds for the order \(d \rho d \theta d \varphi\) are: $$\int_{0}^{\pi / 4} \int_{0}^{2\pi} \int_{0}^{4 \sec\varphi} f(\rho, \varphi, \theta) \rho^{2} \sin \varphi d \rho d \theta d \varphi$$ The new integration bounds for the order \(d \theta d \rho d \varphi\) are: $$\int_{0}^{\pi / 4} \int_{0}^{4 \sec\varphi} \int_{0}^{2\pi} f(\rho, \varphi, \theta) \rho^{2} \sin \varphi d \theta d \rho d \varphi$$

Step-by-step solution

Understand the given bounds

The initial integral is given as: $$\int_{0}^{2 \pi} \int_{0}^{\pi / 4} \int_{0}^{4 \sec\varphi} f(\rho, \varphi, \theta) \rho^{2} \sin \varphi d \rho d \varphi d \theta$$ \(\theta\) varies from \(0\) to \(2\pi\), \(\varphi\) varies from \(0\) to \(\pi/4\), and \(\rho\) varies from \(0\) to \(4\sec\varphi\).

Sketch the region of integration

To better understand how to change the order, sketch the region. Notice that \(\rho\) goes to a maximum of \(4\sec\varphi\). This implies that the outer layer has the shape of a cone with the tip pointing in the positive z-direction with angle \(\pi/4\). Inside this cone, we define the volume for which \(\theta\) varies from \(0\) to \(2\pi\), covering a complete rotation around the z-axis, and \(\varphi\) varying from \(0\) to \(\pi/4\), opening the cone by angle \(\pi/4.\)

Change the order to \(d \rho d \theta d \varphi\)

To change the order of integration to \(d \rho d \theta d \varphi\), first notice that \(\varphi\) still varies from \(0\) to \(\pi/4\) because that bounds the opening angle of the cone. For a fixed \(\varphi,\) \(\theta\) will still vary from \(0\) to \(2\pi\) as it makes a full circle around the z-axis. Lastly, with a fixed \(\varphi\) and \(\theta\), \(\rho\) monotonically increases in magnitude until hitting the outer cone, i.e. \(\rho\) varies from 0 to \(4\sec\varphi.\) Thus, the new order of integration becomes: $$\int_{0}^{\pi / 4} \int_{0}^{2\pi} \int_{0}^{4 \sec\varphi} f(\rho, \varphi, \theta) \rho^{2} \sin \varphi d \rho d \theta d \varphi$$

Change the order to \(d \theta d \rho d \varphi\)

Lastly, for the order \(d \theta d \rho d \varphi\), note that \(\varphi\) still varies from \(0\) to \(\pi/4\) because that represents the opening angle of the cone. With a fixed \(\varphi\), first, the magnitude of \(\rho\) increases until hitting the outer cone, i.e. \(\rho\) varies from 0 to \(4\sec\varphi\). For each fixed \(\rho\) and \(\varphi\), we make a complete turn around the z-axis to cover the whole slice of cone by having \(\theta\) vary from \(0\) to \(2\pi\). Thus, the new order of integration becomes: $$\int_{0}^{\pi / 4} \int_{0}^{4 \sec\varphi} \int_{0}^{2\pi} f(\rho, \varphi, \theta) \rho^{2} \sin \varphi d \theta d \rho d \varphi$$

Final Answer

In conclusion, the orders \(d \rho d \theta d \varphi\) and \(d \theta d \rho d \varphi\) are given by: $$\int_{0}^{\pi / 4} \int_{0}^{2\pi} \int_{0}^{4 \sec\varphi} f(\rho, \varphi, \theta) \rho^{2} \sin \varphi d \rho d \theta d \varphi$$ and $$\int_{0}^{\pi / 4} \int_{0}^{4 \sec\varphi} \int_{0}^{2\pi} f(\rho, \varphi, \theta) \rho^{2} \sin \varphi d \theta d \rho d \varphi$$

Key concepts

Order of Integration

Understanding the order of integration is essential when dealing with multiply integrated functions, especially when working in different coordinate systems such as spherical coordinates. It refers to the sequence in which the multiple integrals are evaluated in iterated integrals. Changing the order of integration might be necessary to simplify computations or to adapt to the geometry of the region being integrated.

In the exercise, the region of integration is a cone with specific bounds for \(\rho\), \(\varphi\), and \(\theta\). When changing the order, we must ensure that the new bounds accurately describe the same region. A correct change of order takes into account how one variable depends on others and adjusts limits accordingly. Careful consideration of the geometry helps to determine the new bounds for each variable, ensuring that all points within the region remain accounted for.

Iterated Integrals

An iterated integral is a way to calculate integrals over a multi-dimensional space by breaking them down into a series of single integrals, each with respect to one variable. The integrals are executed one after another, which is why they are called iterated. It is essential that students grasp this concept to solve problems involving multi-variable functions.

In spherical coordinates, an iterated integral can seem complex due to the extra factors coming from the Jacobian, which includes \(\rho^2\sin\varphi\) in our example. However, the essence is the same: evaluate each integral in the specified sequence, with each step producing a function that is then integrated over the next variable.

Triple Integrals

A triple integral extends the concept of a double integral to three-dimensional space. This type of integral is used to compute quantities like volume, mass, and charge density, where the integrand function describes the density of these quantities throughout a region of space.

In spherical coordinates, every point within the space is represented by \(\rho\), \(\varphi\), and \(\theta\) instead of the cartesian coordinates. Triple integrals in spherical coordinates enable the calculation over regions that are naturally described by spheres, cones, or other radially symmetric shapes, often simplifying the process as compared to Cartesian coordinates.

Volume Integration

The process of volume integration involves using triple integrals to calculate the volume of a three-dimensional region. When working in spherical coordinates, volume integration takes into account the volume element defined by \(\rho^2\sin\varphi d\rho d\varphi d\theta\), which corresponds to a small portion of the spherical shell within the region.

In the given problem, volume integration in spherical coordinates is conducted over a conical region. Understanding the geometric significance of \(\rho\), \(\varphi\), and \(\theta\) in spherical coordinates is crucial to set up these integrals correctly. When students can visualize and understand these elements, volume integration becomes a systematic process of multiplying the density function by the volume element and integrating over the specified region.