Problem 45 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Question: Find the value of the following integral in spherical coordinates: $$\int_{0}^{2\pi} \int_{\pi/6}^{\pi/3} \int_{0}^{2\csc\varphi} \rho^{2}\sin\varphi d \rho d \varphi d \theta$$ Answer: The value of the integral is: $$\int_{0}^{2\pi} \int_{\pi/6}^{\pi/3} \int_{0}^{2\csc\varphi} \rho^{2}\sin\varphi d \rho d \varphi d \theta = \frac{16\pi}{3}+\frac{8\pi\sqrt{3}}{3}$$

Step 1: Transform the integral

We need to transform the Cartesian integral to spherical coordinates with the expression for volume in spherical coordinates, using the Jacobian for the transformation: $$\int_{0}^{2 \pi} \int_{\pi / 6}^{\pi / 3} \int_{0}^{2 \csc \varphi} \rho^{2} \sin \varphi d \rho d \varphi d \theta$$

Step 2: Integrate with respect to ρ

Evaluate the innermost integral with respect to radial distance \(\rho\): $$\int_{0}^{2 \csc \varphi} \rho^2 \sin \varphi d \rho = \left[\frac{1}{3}\rho^3\sin\varphi\right]\bigg|_0^{2\csc\varphi} $$ This results in: $$\left[\frac{1}{3}(2\csc\varphi)^3\sin\varphi\right] = \frac{8}{3}\sin\varphi$$

Step 3: Integrate with respect to 𝜑

Now, we'll evaluate the middle integral with respect to polar angle \(\varphi\): $$\int_{\pi/6}^{\pi/3} \frac{8}{3}\sin\varphi d\varphi = \left[-\frac{8}{3}\cos\varphi\right]\bigg|_{\pi/6}^{\pi/3} $$ This results in: $$-8\left(-\frac{1}{3} - \frac{\sqrt{3}}{6}\right) = \frac{8}{3}+\frac{4\sqrt{3}}{3}$$

Step 4: Integrate with respect to 𝜃

Finally, evaluate the outermost integral with respect to azimuthal angle θ: $$\int_{0}^{2\pi}\left(\frac{8}{3}+\frac{4\sqrt{3}}{3}\right)d\theta=\left[\left(\frac{8}{3}+\frac{4\sqrt{3}}{3}\right)\theta\right]\bigg|_{0}^{2\pi}$$ This results in: $$\left(\frac{8}{3}+\frac{4\sqrt{3}}{3}\right)(2\pi)=\frac{16\pi}{3}+\frac{8\pi\sqrt{3}}{3}$$

Step 5: Final Result

The value of the integral is: $$\int_{0}^{2\pi} \int_{\pi/6}^{\pi/3} \int_{0}^{2\csc\varphi} \rho^{2}\sin\varphi d \rho d \varphi d \theta = \frac{16\pi}{3}+\frac{8\pi\sqrt{3}}{3}$$

Key Concepts

Triple Integral

The triple integral is a powerful mathematical tool that extends the concept of single and double integrals to three dimensions. It allows us to calculate the volume under a surface in three-dimensional space and is particularly useful in physics and engineering for solving problems involving mass, density, and volume. A triple integral is represented by the notation \( \int \int \int f(x, y, z) \, dx \, dy \, dz \), where \( f(x, y, z) \) is a three-variable function, and \( dxdydz \) signifies the infinitesimal volume elements. To perform a triple integral, we usually integrate in the order from the innermost to outermost integral, while the limits of integration represent the region we're integrating over.

When dealing with regions that are not easily describable in Cartesian coordinates, such as spheres and cones, it becomes more efficient to use a coordinate system that matches the symmetry of the problem, like spherical coordinates.

Spherical Integration

Spherical integration is particularly useful when dealing with problems that have spherical symmetry. In such cases, spherical coordinates (\(\rho, \varphi, \theta\)) are employed, where \(\rho\) is the distance from the origin, \(\varphi\) is the polar angle measured from the positive z-axis, and \(\theta\) is the azimuthal angle in the xy-plane from the positive x-axis. Spherical integrals involve integrating over these three variables, accounting for the spherical volume element \( dV = \rho^2 \sin\varphi \, d\rho \, d\varphi \, d\theta \).

To compute a volume in spherical coordinates, one typically evaluates an integral of this form: \( \int_{\theta_1}^{\theta_2} \int_{\varphi_1}^{\varphi_2} \int_{\rho_1}^{\rho_2} f(\rho, \varphi, \theta) \rho^2 \sin\varphi \, d\rho \, d\varphi \, d\theta \). The limits of integration \(\theta_1, \theta_2, \varphi_1, \varphi_2, \rho_1, \rho_2\) define the spherical region of integration.

Volume in Spherical Coordinates

Calculating the volume in spherical coordinates involves using the spherical volume element \( dV = \rho^2 \sin\varphi \, d\rho \, d\varphi \, d\theta \), which arises from the geometry of the spherical coordinate system. The volume element represents a small 'chunk' of space, corresponding to a wedge-shaped segment of a sphere.

To find the volume of a spherical region, we integrate the volume element over the desired region. For example, the formula for the volume of a sphere of radius R is obtained by integrating the volume element from \(0\) to \(R\) for \(\rho\), from \(0\) to \(\pi\) for \(\varphi\), and from \(0\) to \(2\pi\) for \(\theta\): \( V = \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{R} \rho^2 \sin\varphi \, d\rho \, d\varphi \, d\theta \). This results in the familiar \(\frac{4}{3}\pi R^3\), the volume of a sphere.

Jacobian

The Jacobian is a determinant that arises in multivariable calculus, specifically in transformations between coordinate systems. It's used to scale the volume when changing from one coordinate system to another, such as from Cartesian to spherical coordinates. Mathematically, the Jacobian for spherical coordinates \(J(\rho, \varphi, \theta)\) is \( \rho^2 \sin\varphi \), and it represents how much the volume element \(dV\) changes when transforming between these systems.

The Jacobian is crucial because it ensures the preservation of the proper scale of volume, area, or length during integration. For instance, when computing integrals in spherical coordinates, we must multiply the integrand by the Jacobian to account for the distortion in volume elements due to the coordinate change. This is precisely why the volume element in spherical coordinates includes the Jacobian factor \( \rho^2 \sin\varphi \).