Problem 41 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Question: Evaluate the triple integral of the function \(\frac{1}{(x^2+y^2+z^2)^{3/2}}\) over the region \(D\), where \(D\) is the solid between the 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 1: Transform the function from Cartesian to spherical coordinates

To transform the integral from Cartesian to spherical coordinates, we will use the following transformations: \[ x = \rho \sin\phi \cos\theta \\ y = \rho \sin\phi \sin\theta \\ z = \rho \cos\phi \] Additionally, we know that the differential volume element \(dV\) in spherical coordinates is given by \(dV = \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta\). Then, the denominator in the function becomes \(\rho^3\), and thus the function transforms to: \[ \frac{1}{\rho^3} \] This gives us the new integral, \[ \iiint_D \frac{\rho^2 \sin\phi \, d\rho \, d\phi \, d\theta}{\rho^3} \]

Step 2: Identify the limits of integration for spherical coordinates

Since the integral is over the solid between the spheres of radius 1 and 2, we have the following limits for each spherical coordinate: - For \(\rho\), the radius ranges from 1 to 2, since the solid is between two spheres with radii 1 and 2, so \(1 \leq \rho \leq 2\). - For \(\phi\), the polar angle ranges from 0 to \(\pi\), since the solid covers the entire sphere in the \(xy\)-plane, so \(0 \leq \phi \leq \pi\). - For \(\theta\), the azimuthal angle ranges from 0 to \(2\pi\), since the solid covers the entire circumference of the sphere in the \(xy\)-plane, so \(0 \leq \theta \leq 2\pi\). Now we have the new limits of integration.

Step 3: Evaluate the triple integral

The integral now looks like this: \[ \int_{0}^{2\pi} \int_{0}^{\pi} \int_{1}^{2} \frac{\rho^2 \sin\phi}{\rho^3} \, d\rho \, d\phi \, d\theta \] First, we cancel out a \(\rho\) term in the numerator and the denominator, simplifying the integrand: \[ \int_{0}^{2\pi} \int_{0}^{\pi} \int_{1}^{2} \frac{\sin\phi}{\rho} \, d\rho \, d\phi \, d\theta \] Now, we evaluate the triple integral: \[ \int_{0}^{2\pi} d\theta \int_{0}^{\pi} \sin\phi \, d\phi \int_{1}^{2} \frac{1}{\rho} \, d\rho \] Evaluate each integral separately: \[ \int_{0}^{2\pi} d\theta = 2\pi \\ \int_{0}^{\pi} \sin\phi \, d\phi = [-\cos \phi]_{0}^{\pi} = 2 \\ \int_{1}^{2} \frac{1}{\rho} \, d\rho = [\ln \rho]_{1}^{2} = \ln 2 \] Now, multiply the three evaluated integrals together to obtain the final result: \[ (2\pi)(2)(\ln 2) = 4\pi\ln 2 \] Therefore, the value of the given integral in spherical coordinates is \(4\pi\ln 2\).