Question

Evaluate the following integrals in spherical coordinates. \(\iiint_{D} \frac{1}{\left(x^{2}+y^{2}+z^{2}\right)^{3 / 2}} d V ; D\) is the region between the spheres of radius 1 and 2 centered at the origin.

Short answer

Question: Evaluate the triple integral of the function \(\frac{1}{\left(x^{2}+y^{2}+z^{2}\right)^{3 / 2}}\) over the region bounded by two spheres of radius 1 and 2 centered at the origin, using spherical coordinates. Answer: The value of the triple integral is \(4\pi\ln(2)\).

Step-by-step solution

Convert to spherical coordinates

First, we rewrite the function in spherical coordinates. Recall that in spherical coordinates, we have: \(x = \rho\sin\phi\cos\theta, y = \rho\sin\phi\sin\theta, z = \rho\cos\phi\) and \(dV = \rho^2\sin\phi d\rho d\phi d\theta\). Now rewrite the given function in spherical coordinates: \(\frac{1}{\left(x^{2}+y^{2}+z^{2}\right)^{3 / 2}} = \frac{1}{(\rho^2\sin^2\phi\cos^2\theta + \rho^2\sin^2\phi\sin^2\theta + \rho^2\cos^2\phi)^{3/2}} = \frac{1}{(\rho^2)^{3/2}} = \frac{1}{\rho^3}\).

Set up the limits of integration and integrate

The region D is bounded by two spheres of radius 1 and 2 centered at the origin. Now we determine the limits of integration for \(\rho\), \(\phi\), and \(\theta\): - \(\rho\) varies from 1 to 2, since the region D is between the spheres of radius 1 and 2. - \(\phi\) varies from 0 to \(\pi\), as it spans the entire region from north to south pole. - \(\theta\) varies from 0 to \(2\pi\), as it goes around the entire sphere. Now, integrate using these limits of integration: \(\iiint_{D} \frac{1}{\left(x^{2}+y^{2}+z^{2}\right)^{3 / 2}} d V = \int_{1}^{2}\int_{0}^{\pi}\int_{0}^{2\pi}\frac{\rho^2\sin\phi}{\rho^3}d\theta d\phi d\rho\)

Evaluate the integral

Now, we can integrate with respect to \(\theta\), \(\phi\), and \(\rho\) one variable at a time: \(\int_{1}^{2}\int_{0}^{\pi}\int_{0}^{2\pi}\frac{\rho^2\sin\phi}{\rho^3}d\theta d\phi d\rho = \int_{1}^{2}\int_{0}^{\pi}\left[\frac{\rho\sin\phi}{\rho^2}\right]\left[\int_{0}^{2\pi}d\theta\right]d\phi d\rho = \int_{1}^{2}\int_{0}^{\pi}\frac{\rho\sin\phi}{\rho^2}(2\pi - 0)d\phi d\rho\) Next, integrate with respect to \(\phi\): \(\int_{1}^{2}\int_{0}^{\pi}\frac{2\pi\rho\sin\phi}{\rho^2}d\phi d\rho = 2\pi\int_{1}^{2}\left(\frac{1}{\rho}\right)\left[-\cos\phi\right]_{0}^{\pi}\, d\rho = 2\pi\int_{1}^{2}\frac{\cos(0) - \cos(\pi)}{\rho}d\rho\) Finally, integrate with respect to \(\rho\): \(2\pi\int_{1}^{2}\frac{2}{\rho}d\rho = 4\pi\int_{1}^{2}\frac{1}{\rho}d\rho = 4\pi\left[\ln|\rho|\right]_{1}^{2}\)

Calculate the final result

Now, evaluate the result: \(4\pi\left[\ln|\rho|\right]_{1}^{2} = 4\pi(\ln(2) - \ln(1)) = 4\pi\ln(2)\) So, the value of the triple integral is \(4\pi\ln(2)\).

Key concepts

triple integrals

Triple integrals are a powerful tool used in calculus to find the volume under a three-dimensional surface. Think of them as an extension of double integrals, which are used to calculate areas. In this exercise, the triple integral helps us to evaluate complex volumes in three-dimensional spaces defined by certain functions.

The calculation involves integrating a function with respect to three variables, usually represented as \(x\), \(y\), and \(z\). When dealing with triple integrals in \(xyz\) space, one must often transform into other coordinate systems, such as spherical coordinates, for simpler calculations.

Here is a simple step-by-step process to understand triple integrals better:

• Identify the function to be integrated over the three-dimensional region.
• Set the limits for each variable according to the defined region.
• Integrate the function stepwise with respect to each variable in succession.

Remember, successfully evaluating a triple integral hinges on setting correct limits and systematically evaluating the integral for each variable.

spherical integration

Spherical integration, as the name suggests, involves using spherical coordinates. It's most frequently used when dealing with volumes of spheres or spherical shapes. This coordinate system is particularly useful when symmetry around a central point, like the origin, is present.

In the spherical coordinate system, any point in three-dimensional space is determined by three parameters: \(\rho\), \(\phi\), and \(\theta\). Here's a bit more detail:

• \(\rho\) represents the distance from the origin to the point.
• \(\phi\) is the angle measured from the positive \(z\)-axis, ranging from 0 to \(\pi\).
• \(\theta\) is the angle in the \(xy\)-plane from the positive \(x\)-axis, ranging from 0 to \(2\pi\).

To convert a triple integral from Cartesian to spherical coordinates, replace the Cartesian variables with their spherical equivalents and adjust the differential volume accordingly: \(dV = \rho^2\sin\phi d\rho d\phi d\theta\). This equation accounts for the changes in space as you move from one coordinate system to the other. Spherical integrals make complex calculations neater, especially when applied to spherical regions.

volume integrals

Volume integrals are essential for calculating the capacity or space within a three-dimensional shape. They generalize the two-dimensional area concept extended into the third dimension.

Volume integrals can be set up either in Cartesian, cylindrical, or spherical coordinates, depending on the shape of the region and the symmetry present. In this exercise, spherical coordinates simplify the process as they align more naturally with spherical objects.

Here’s a quick summary of how to use volume integrals effectively:

• Choose the most fitting coordinate system based on the symmetry of the object.
• Determine the limits of integration carefully, as they define the boundaries of the region.
• Follow the order of integration from the innermost variable to the outermost.
• Evaluate the integral iteratively, starting inside and moving outward.

By carefully selecting the coordinate system and setting accurate limits, volume integrals offer a structured approach to calculating volumes, even in complex regions.