Question

Evaluate the following integrals in spherical coordinates. \(\int_{0}^{2 \pi} \int_{\pi / 6}^{\pi / 3} \int_{0}^{2 \csc \varphi} \rho^{2} \sin \varphi d \rho d \varphi d \theta\)

Short answer

Question: Evaluate the integral of the given function in spherical coordinates with the given limits of integration. Integrand: \(\rho^2 \sin\varphi\) Limits of integration: 1. For theta: from 0 to \(2\pi\) 2. For phi: from \(\frac{\pi}{6}\) to \(\frac{\pi}{3}\) 3. For rho: from 0 to \(2\csc\varphi\) Answer: \(\frac{8\pi}{3}\)

Step-by-step solution

Integration with respect to rho

Integrate the integrand \(\rho^2 \sin\varphi\) with respect to \(\rho\) by keeping \(\varphi\) and \(\theta\) constant from \(0\) to \(2\csc\varphi\). The integral is: $$\int_{0}^{2\csc\varphi} \rho^{2} \sin \varphi d\rho$$ Using the power rule of integration, we get: $$[\frac{1}{3}\rho^{3}\sin\varphi]_{0}^{2\csc\varphi} = \frac{8}{3}\sin\varphi$$ Now, the integral becomes: $$\int_{0}^{2\pi} \int_{\pi / 6}^{\pi / 3} \frac{8}{3}\sin\varphi d\varphi d\theta$$

Integration with respect to phi

Now, integrate the result from Step 1 with respect to \(\varphi\) from \(\frac{\pi}{6}\) to \(\frac{\pi}{3}\) by keeping \(\theta\) constant. The integral is: $$\int_{\pi/6}^{\pi/3} \frac{8}{3}\sin\varphi d\varphi$$ Integrating \(\sin\varphi\) with respect to \(\varphi\), we get: $$[\frac{8}{3}(-\cos\varphi)]_{\pi/6}^{\pi/3} = \frac{8}{3}(-\cos(\pi/3) + \cos(\pi/6)) = \frac{4}{3}$$ Now, the integral becomes: $$\int_{0}^{2\pi} \frac{4}{3} d\theta$$

Integration with respect to theta

Finally, integrate the result from Step 2 with respect to \(\theta\) from \(0\) to \(2\pi\). The integral is: $$\int_{0}^{2\pi} \frac{4}{3} d\theta$$ Integrating the constant, we get: $$[\frac{4}{3}\theta]_{0}^{2\pi} = \frac{4}{3}\times 2\pi$$

Final Answer

The value of the given integral in spherical coordinates is: $$\boxed{\frac{8\pi}{3}}$$

Key concepts

Integration in Spherical Coordinates

Let's break down the integration process in spherical coordinates, which is a coordinate system that is very useful in solving problems involving symmetry around a point, such as those involving spheres. When working in spherical coordinates for integration, we typically have integrals in the form of \( \iiint f(\rho, \varphi, \theta) \rho^2 \sin \varphi \, d\rho \, d\varphi \, d\theta \). The volume element \( dV \) in spherical coordinates is \( \rho^2 \sin \varphi \, d\rho \, d\varphi \, d\theta \), which accounts for the three-dimensional nature of the problem.

The first step in solving the given integral involves integration with respect to \( \rho \), the radial distance from the origin. The bounds for \( \rho \) are typically determined by the geometric properties of the problem, such as the radius of a sphere. After integrating with respect to \( \rho \), the expression simplifies, allowing us to then integrate with respect to \( \varphi \), and finally, \( \theta \). This multi-step process is essential for solving the integral and finding the volume or area integral that the problem describes.

Calculus and Its Role in Integration

Calculus is the branch of mathematics that allows us to analyze changes and motion using tools such as derivatives and integrals. Integration, one of the core concepts of calculus, is particularly useful when calculating areas, volumes, and other quantities where standard geometry does not apply.

In this exercise, the integration process is focused on calculating an integral using spherical coordinates, which is a clear example of how calculus can simplify solving complex geometric problems. We start by integrating a function over a region to find volume using triple integrals. Each step of the integration process—first \( \rho \), then \( \varphi \), and finally \( \theta \)—relates directly back to fundamental calculus concepts. These concepts are built on the power rule and the known antiderivatives of trigonometric functions, which are crucial for solving these integrals accurately. The systematic step-by-step solving of integrals aids in finding precise solutions to otherwise complex physical and mathematical phenomena.

Understanding Triple Integrals

A triple integral, as used in this problem, extends the idea of a definite integral to functions of three variables and is often used for computing volumes. Unlike single or double integrals, a triple integral allows integration over three-dimensional regions. In spherical coordinates, this means integrating a function over a spherical volume or region defined in three-dimensional space.

The process involves integrating one variable at a time, each within its own specified bounds. For the given problem, the bounds for \( \rho \) were from \( 0 \) to \( 2\csc \varphi \), for \( \varphi \) were from \( \frac{\pi}{6} \) to \( \frac{\pi}{3} \), and for \( \theta \) were from \( 0 \) to \( 2\pi \). Each bound is defined based on the region we are interested in within this spherical coordinate system. Triple integrals, such as this one, enable us to compute the volume and other characteristics of the region encompassed by these limits, often simplifying the computation of complex problems in physics and engineering.