Problem 81 - Advanced Engineering Mathematics

Question

Find the mass of the solid that is bounded above by the hemisphere \(z=\sqrt{25-x^{2}-y^{2}}\) and below by the plane \(z=4\) if the density at a point \(P\) is inversely proportional to the distance from the origin. [Hint: Express the upper \(\phi\) limit of integration as an inverse cosine.]

Step-by-Step Solution

Answer

Mass is calculated as a definite integral involving density.

Step 1: Understand the Region

The solid is bounded above by the hemisphere \( z = \sqrt{25 - x^2 - y^2} \) and below by the plane \( z = 4 \). The hemisphere is a part of a sphere with radius 5 centered at the origin. The sphere above the plane starts at \( z = 4 \).

Step 2: Set Up Cylindrical Coordinates

In cylindrical coordinates, \( x = r\cos\theta \), \( y = r\sin\theta \), and \( z = z \). The hemisphere then translates to \( z = \sqrt{25 - r^2} \). The projection on the xy-plane is a circle with radius 3 centered at the origin.

Step 3: Establish the Density Function

The density function, \( \rho(x, y, z) \), is inversely proportional to the distance from the origin: \( \rho = \frac{k}{\sqrt{x^2 + y^2 + z^2}} \). In cylindrical coordinates, this becomes \( \rho = \frac{k}{\sqrt{r^2 + z^2}} \).

Step 4: Define Integration Limits

We will integrate \( r \) from 0 to 3, \( \theta \) from 0 to \( 2\pi \), and \( z \) from 4 to the hemisphere boundary \( \sqrt{25 - r^2} \).

Step 5: Determine the Mass Integral

The mass \( M \) of the solid is given by the triple integral:\[ M = \int_{0}^{2\pi} \int_{0}^{3} \int_{4}^{\sqrt{25 - r^2}} \frac{k \cdot r}{\sqrt{r^2 + z^2}} \, dz \, dr \, d\theta\]This is because the density function is multiplied by \( r \) to account for the transformation to cylindrical coordinates.

Step 6: Evaluate the Integral

First integrate with respect to \( z \):\[ \int_{4}^{\sqrt{25 - r^2}} \frac{r}{\sqrt{r^2 + z^2}} \, dz = \left[ -\sqrt{r^2 + z^2} \right]_{z=4}^{z=\sqrt{25 - r^2}}\]This evaluates to:\[ -\sqrt{25} + \sqrt{r^2 + 16} = -5 + \sqrt{r^2 + 16}\]

Step 7: Integrate Over \( r \)

Next integrate with respect to \( r \):\[ \int_{0}^{3} [-5 + \sqrt{r^2 + 16}] \, r \, dr\]This needs to be computed and includes a substitution for the second term.

Step 8: Substitution and Solution

Use the substitution \( u = r^2 + 16 \), \( du = 2r \, dr \), to handle the square root term. Substitute and integrate to find the contribution from this term.

Step 9: Integrate Over \( \theta \)

Finally, multiply the result by the \( \theta \) integration over \( 0 \) to \( 2\pi \) to account for circular symmetry:\[ M = 2\pi\left[ \text{Result from } \int_{0}^{3} [-5r + \sqrt{r^2 + 16}r] \, dr \right]k \]

Step 10: Solve and Simplify

Combine all evaluated parts to find the total mass of the solid. Ensure to include the constant \( k \) when expressing the final answer.

Key Concepts

Triple Integral

A triple integral is a powerful tool in calculus, particularly useful for determining quantities such as volume, mass, and center of mass in three-dimensional regions. For this exercise, a triple integral is used to find the mass of the solid bounded between a hemisphere and a plane. The solid is described using cylindrical coordinates, providing a straightforward framework to write the limits of integration. In general, the triple integral for mass is defined as

• \( \int \int \int \rho(x, y, z) \, dV \)

where \( \rho(x, y, z) \) represents the density function and \( dV \) is the volume element represented in the chosen coordinate system, in this case, cylindrical coordinates. By integrating over the specified limits for radius, angle, and height, we can capture the mass of the solid in its entirety.

Cylindrical Coordinates

Cylindrical coordinates provide an alternative to Cartesian coordinates, particularly advantageous when working with rotationally symmetric objects like spheres and cylinders. In cylindrical coordinates, any point in space is represented by the combination of

• \( r \) - the radius or radial distance from the z-axis
• \( \theta \) - the angle with the positive x-axis
• \( z \) - the height above the xy-plane

These coordinates simplify the integration process compared to Cartesian coordinates, as they adapt to the symmetry of the problem. In this exercise, they are used to translate a hemisphere boundary into a workable form, where the original spherical boundary becomes \( z = \sqrt{25 - r^2} \), and the projection on the xy-plane forms a circular region with radius 3.

Density Function

Density functions allow for variable distribution of mass throughout a volume, which can depend on the position within the space. In this problem, the density function is inversely proportional to the distance from the origin, expressed mathematically as

• \( \rho(x, y, z) = \frac{k}{\sqrt{x^2 + y^2 + z^2}} \)
• In cylindrical coordinates, it simplifies to \( \rho(r, \theta, z) = \frac{k}{\sqrt{r^2 + z^2}} \)

Here, the constant \( k \) could be any real number, often given in problems involving density. This means that the contribution to the mass from any point decreases as it moves further from the origin. Hence, locations closer to the origin contribute more to the total mass.

Sphere Integration

Sphere integration involves solving integrals over spherical geometries, which often involves switching to spherical or cylindrical coordinates for simplification. Here, by dividing the integration region into a cylinder below a hemisphere, calculations become more tractable. The boundaries are specified by the hemisphere \( z = \sqrt{25 - r^2} \) and the plane \( z = 4 \). This setup implies integrating first over \( z \), then \( r \), finally \( \theta \), covering:

• The z-bound, height, varies from the plane to the hemisphere above.
• The radial coordinate \( r \) limits extend from the origin to the edge of the base circle.
• The angle \( \theta \) completes the circle by spanning from 0 to \( 2\pi \).

This way, the integral accounts for the complete geometry of the provided boundaries, crucial for accurately assessing volumes or other properties such as mass in spherical contexts.