Question

A solid is described along with its density function. Find the mass of the solid using spherical coordinates. The spherical shell bounded between \(x^{2}+y^{2}+z^{2}=16\) and \(x^{2}+y^{2}+z^{2}=25\) with density function \(\delta(x, y, z)=\) \(\sqrt{x^{2}+y^{2}+z^{2}}\).

Short answer

The mass of the solid spherical shell is \(369\pi\).

Step-by-step solution

Convert to Spherical Coordinates

In spherical coordinates, the relations are: \( x = \rho \sin\phi \cos\theta \), \( y = \rho \sin\phi \sin\theta \), \( z = \rho \cos\phi \). Therefore, \(x^2 + y^2 + z^2 = \rho^2\). The density function becomes \(\delta(\rho) = \rho\). We also know that the shell is bounded by \(\rho = 4\) and \(\rho = 5\).

Set Up the Integral

The mass \(M\) of the solid is given by the integral of the density function over the volume of the shell. In spherical coordinates, the volume element is \(dV = \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta\). Therefore, the integral to find the mass is:\[M = \int_0^{2\pi} \int_0^{\pi} \int_4^5 \rho^3 \sin\phi \, d\rho \, d\phi \, d\theta\]

Integrate with Respect to \(\rho\)

Integrate the innermost integral with respect to \(\rho\):\[ \int_4^5 \rho^3 \, d\rho = \left[ \frac{\rho^4}{4} \right]_4^5 = \left( \frac{5^4}{4} - \frac{4^4}{4} \right) = \frac{625}{4} - \frac{256}{4} = \frac{369}{4} \]

Integrate with Respect to \(\phi\)

Now integrate with respect to \(\phi\):\[ \int_0^{\pi} \left( \frac{369}{4} \right) \sin\phi \, d\phi = \frac{369}{4} \left[ -\cos\phi \right]_0^{\pi} = \frac{369}{4} (2) = \frac{738}{4} = \frac{369}{2} \]

Integrate with Respect to \(\theta\)

Finally, integrate with respect to \(\theta\):\[ \int_0^{2\pi} \frac{369}{2} \, d\theta = \frac{369}{2} \left[ \theta \right]_0^{2\pi} = \frac{369}{2} (2\pi) = 369\pi \]

Conclusion and Final Answer

The mass of the solid spherical shell is the result of the last integration, which gives:\[ M = 369\pi \]

Key concepts

Mass of a Solid

When we talk about finding the mass of a solid object, we're essentially looking to determine how much matter is contained within it. This gets a bit more complicated when the solid does not have a uniform density throughout its volume. Instead of multiplying a constant density by the volume, we use calculus to integrate over the solid's volume.
For example, consider a solid spherical shell like the one in our exercise. This object exists in three dimensions and has mass distributed based on the density function provided. The density function can vary with position, which is common in problems involving real-world materials. To find the mass of such a solid, one must understand how to process and utilize a given density function effectively.

Density Function

A density function describes how mass is distributed throughout an object. In spherical coordinates, this function is often more intuitive to use for bodies exhibiting spherical symmetry. In our example, the density function is given by \( \delta(x, y, z) = \sqrt{x^2 + y^2 + z^2} \),which simplifies to \( \delta(\rho) = \rho \) in spherical coordinates, where \( \rho \) is the radial distance from the origin.
This function indicates that the density increases with distance from the center of the sphere. This increasing density with radius could represent a physical scenario where a material becomes denser the further you move from its center.
Understanding a density function is crucial for problems involving variable density, as it directly affects calculations of mass and center of mass.

Triple Integrals

Triple integrals allow us to calculate volumes, masses, and other properties of three-dimensional objects. They are particularly useful when dealing with objects that have varying properties like density.
In the context of mass calculations, the triple integral is set up based on the volume confines and the density function. For the spherical shell described, using spherical coordinates simplifies the integration due to symmetry. The integral is expressed as: \[M = \int_0^{2\pi} \int_0^{\pi} \int_{\rho_1}^{\rho_2} \rho^3 \sin\phi \, d\rho \, d\phi \, d\theta\] where \( \rho_1 \) and \( \rho_2 \) are the inner and outer radii of the shell, and \( \sin\phi \) accounts for the angular distribution in the sphere.
Triple integrals help us consolidate information across multiple dimensions, making them a cornerstone in multivariable calculus and invaluable for applications in physics and engineering.