Question

A solid is described along with its density function. Find the mass of the solid using cylindrical coordinates. Bounded by the cylinder \(x^{2}+y^{2}=4\) and the planes \(z=0\) and \(z=4\) with density function \(\delta(x, y, z)=\sqrt{x^{2}+y^{2}}+1\).

Short answer

The mass of the solid is \( \frac{112\pi}{3} \).

Step-by-step solution

Convert to Cylindrical Coordinates

In cylindrical coordinates, we have the relationships: - The radial distance, \(r\), such that \(x^2 + y^2 = r^2\).- The angle \(\theta\), which is the angle from the positive x-axis.- The height, \(z\), which remains the same.For the given cylinder \(x^2 + y^2 = 4\), this translates to \(r = 2\). The planes \(z = 0\) and \(z = 4\) give the vertical height bounds.

Set Up the Integral for Mass

The mass of the solid can be calculated as the triple integral of the density function over the volume. The density function in cylindrical coordinates becomes \(\delta(r, \theta, z) = r + 1\). We need to integrate this function over the region described by: - \(0 \leq r \leq 2\)- \(0 \leq \theta \leq 2\pi\)- \(0 \leq z \leq 4\)Therefore, the integral becomes:\[ M = \int_{0}^{2\pi} \int_{0}^{2} \int_{0}^{4} (r + 1) \, r \, dz \, dr \, d\theta \]

Evaluate the Integral with Respect to z

First, integrate with respect to \(z\):\[ \int_{0}^{4} (r + 1) \, dz = (r + 1)z \bigg|_{0}^{4} = 4(r + 1) \]This reduces the integral to:\[ M = \int_{0}^{2\pi} \int_{0}^{2} 4(r + 1) \, r \, dr \, d\theta \]

Evaluate the Integral with Respect to r

Integrate with respect to \(r\):\[ \int_{0}^{2} 4(r^2 + r) \, dr = 4 \left( \frac{r^3}{3} + \frac{r^2}{2} \right) \bigg|_{0}^{2} = 4 \left( \frac{8}{3} + 2 \right) = 4 \left( \frac{14}{3} \right) = \frac{56}{3} \]This simplifies the integral to:\[ M = \int_{0}^{2\pi} \frac{56}{3} \, d\theta \]

Evaluate the Integral with Respect to \(\theta\)

Finally, integrate with respect to \(\theta\):\[ \int_{0}^{2\pi} \frac{56}{3} \, d\theta = \frac{56}{3} \cdot \theta \bigg|_{0}^{2\pi} = \frac{56}{3} \cdot 2\pi = \frac{112\pi}{3} \]This gives us the mass of the solid.

Key concepts

Density Function

In mathematics, a density function is an important concept used to describe how mass is distributed within a given volume. For a solid object, the density function \( \delta(x, y, z) \) provides the density at any point within the object. When working with problems involving density, the density function can vary depending on the position within the object. For example, in this particular problem, the density function is given as \( \delta(x, y, z) = \sqrt{x^{2}+y^{2}}+1 \).
This means that the density increases with the radial distance from the z-axis within the cylindrical coordinates. Understanding the density function is crucial when calculating the mass of an object, as it allows us to set up an integral that accounts for variations in density across the volume.

Triple Integral

A triple integral is a method used in calculus to compute the volume under a surface within a three-dimensional space. It is akin to an 'advanced' version of double integrals but extends the calculations into the third dimension.
The usage of triple integrals becomes particularly handy when working with complex shapes and functions, such as those encountered in density calculations.To find the mass of a solid with a varying density, you integrate the density function throughout the entire volume of the solid. The problem at hand requires us to compute the mass using the triple integral set up in cylindrical coordinates.
The prepared integral for mass is:\[ M = \int_{0}^{2\pi} \int_{0}^{2} \int_{0}^{4} (r + 1) \, r \, dz \, dr \, d\theta \]This integral is solved in layers: starting with the innermost, and then moving outwards. This ensures each dimension (z, r, and \( \theta \)) is properly accounted for.

Mass Calculation

Mass calculation using integration involves taking the density at each tiny piece of the volume and summing everything together. This is exactly what a triple integral does.For computing the mass of the solid, we set up an integral of the form:\[ M = \int (\text{density function}) \, dV \]In this scenario, we first convert the density function into cylindrical coordinates: \( \delta(r, \theta, z) = r + 1 \). Then, we integrate this function over the specified region. The intervals for the variables \( r \), \( \theta \), and \( z \) are decided based on the geometry of the problem. The steps include:

• Integrating with respect to \( z \) first, as it has fixed boundaries of 0 and 4.
• Followed by \( r \), from 0 to 2, defined by the cylinder's radius.
• And finally, \( \theta \) from 0 to \( 2\pi \) covering the full circle around the axis.

The final result, \( \frac{112\pi}{3} \), signifies the accumulated mass of the solid.

Calculus Problem Solving

Solving calculus problems, particularly those involving three-dimensional shapes and density functions, requires a systematic approach.
Cylindrical coordinates are specifically chosen for problems like this due to their inherent symmetry with certain shapes, such as cylinders. Identifying the boundaries and translating them into appropriate limits for integration is key to setting up the problem properly.

Steps to Approach the Problem:

• Understand the geometry: Recognize the shape in question and its characteristics. This helps in identifying the coordinate system to use.
• Transform the function: Convert the given density function to match the chosen coordinate system.
• Set up the integral: Define the limits of integration based on the geometrical constraints.
• Evaluate step-by-step: Solve the integral by sequentially integrating over each variable, starting from the innermost integral.
• Verify the solution: Check the solution against the physical intuition of the problem for validation.

This structured approach allows for efficient and accurate problem-solving in calculus, particularly when working with complex integrals and three-dimensional spaces.