91Ó°ÊÓ

First, let's transform the given integral from Cartesian to spherical coordinates using the following transformations: $$ x = \rho \sin{\phi} \cos{\theta},\\ y = \rho \sin{\phi} \sin{\theta},\\ z = \rho \cos{\phi}, $$ where \(\rho\) is the distance from the origin, \(\phi\) is the polar angle (the angle between the positive \(z\)-axis and the line connecting the point and the origin), and \(\theta\) is the azimuthal angle (the angle between the positive \(x\)-axis and the projection of the point onto the \(x\)-\(y\) plane). Now, let's find the Jacobian for this transformation: $$ J = \frac{\partial(x, y, z)}{\partial(\rho, \phi, \theta)} = \begin{bmatrix} \sin{\phi} \cos{\theta} & \rho \cos{\phi} \cos{\theta} & -\rho \sin{\phi} \sin{\theta} \\ \sin{\phi} \sin{\theta} & \rho \cos{\phi} \sin{\theta} & \rho \sin{\phi} \cos{\theta} \\ \cos{\phi} & -\rho \sin{\phi} & 0 \end{bmatrix} $$ The determinant of this matrix is: $$ J = \rho^2 \sin{\phi} $$

Now, we will express the integral in spherical coordinates and use the Jacobian we found above: $$ \int_{D} \sqrt{x^2+y^2+z^2} \mathrm{~d}(x, y, z) = \int\int\int_{D} \rho \cdot \rho^2 \sin{\phi} \mathrm{~d}\rho \mathrm{~d}\phi \mathrm{~d}\theta $$ The limits of integration will be: - For \(\rho\): \(r\) to \(R\) to take care of the spherical shell, - For \(\phi\): \(0\) to \(\pi\), - For \(\theta\): \(0\) to \(2\pi\) Thus, the integral becomes: $$ \int_{r}^{R} \int_{0}^{\pi} \int_{0}^{2\pi} \rho^3 \sin{\phi} \mathrm{~d}\theta \mathrm{~d}\phi \mathrm{~d}\rho $$

We can now evaluate the integral step by step: 1. Integrating with respect to \(\theta\): $$ \int_{0}^{2\pi} \mathrm{~d}\theta = 2\pi $$ 2. Integrating with respect to \(\phi\): $$ \int_{0}^{\pi} \sin{\phi} \mathrm{~d}\phi = [-\cos{\phi}]_{0}^{\pi} = (1 - 0) + (0 - (-1)) = 2 $$ 3. Integrating with respect to \(\rho\): $$ \int_{r}^{R} \rho^3 \mathrm{~d}\rho = \frac{1}{4}[\rho^4]_{r}^{R} = \frac{1}{4}(R^4 - r^4) $$ 4. Finally, combine all the results: $$ \int_{D} \sqrt{x^2+y^2+z^2} \mathrm{~d}(x, y, z) = 2\pi \cdot 2 \cdot \frac{1}{4}(R^4 - r^4) = \pi(R^4 - r^4) $$ Thus, the value of the integral over the spherical shell \(D\) is: $$ \int_{D} \sqrt{x^2+y^2+z^2} \mathrm{~d}(x, y, z) = \pi(R^4 - r^4) $$