Problem 73 - Calculus Early Transcendentals

Question

Density distribution A right circular cylinder with height \(8 \mathrm{cm}\) and radius \(2 \mathrm{cm}\) is filled with water. A heated filament running along its axis produces a variable density in the water given by \(\rho(r)=1-0.05 e^{-0.01 r^{2}} \mathrm{g} / \mathrm{cm}^{3}(\rho\) stands for density here, not the radial spherical coordinate). Find the mass of the water in the cylinder. Neglect the volume of the filament.

Step-by-Step Solution

Answer

Answer: The mass of the water in the cylinder is approximately \(M \approx -20\pi e^{-0.04} + 32\pi \mathrm{g}\).

Step 1: Volume Integration Limits

For r, the integration limits will be from 0 to 2 cm (the radius of the cylinder). For z, the limits will be from 0 to 8 cm (the height of the cylinder). For θ, the limits will be from 0 to 2π (the total angle around the cylinder).

Step 2: Integrate the Density Function

Let's integrate the density function with respect to r, z, and θ: \(\int_{0}^{8} \int_{0}^{2 \pi} \int_{0}^{2} \rho(r)r\, dr\, d\theta \, dz\)

Step 3: Perform Inner Integration (r)

First, we will integrate with respect to r: \(\int_{0}^{2}(1 - 0.05 e^{-0.01r^2}r\,dr\) Now we perform integration by parts with \(u=r\) and \(dv=(1 - 0.05 e^{-0.01r^2})dr\). Now we'll get: \([rv - 0.05 \int re^{-0.01r^2} dr]_0^2\) The remaining integral can be solved with substitution method. Let \(t = -0.01r^2\), then \(dt=-0.02r\,dr\), so \(-0.05\int re^{-0.01r^2} dr=-\frac{5}{2}\int e^{t} dt\). Now, substitute back: \([rv - 0.05 \int re^{-0.01r^2} dr]_0^2 = [2v - \frac{5}{2}(e^{t})]_0^2\).

Step 4: Perform Middle Integration (θ)

Next, we will integrate with respect to θ: \(\int_{0}^{2\pi} ([2v - \frac{5}{2}(e^{t})]_0^2)d\theta\) Simply multiply by (2π) as there are no θ terms: \(2\pi([2v - \frac{5}{2}(e^{t})]_0^2)\).

Step 5: Perform Outer Integration (z)

Finally, we will integrate with respect to z: \(\int_{0}^{8} 2\pi([2v - \frac{5}{2}(e^{t})]_0^2) dz\) Simply multiply by 8 since there are no z terms: \(8\cdot2\pi([2v - \frac{5}{2}(e^{t})]_0^2)\).

Step 6: Compute the Mass of the Water

Now that we have integrated over the volume of the cylinder, we can calculate the mass of the water: \(M = 8\cdot2\pi([2v - \frac{5}{2}(e^{t})]_0^2) = -20\pi e^{-0.04} + 32\pi\) Thus, the mass of the water in the cylinder is approximately \(M \approx -20\pi e^{-0.04} + 32\pi \mathrm{g}\).

Key Concepts

Volume Integration

Volume integration is a mathematical technique used to calculate the volume of a three-dimensional object by integrating a function over a given region. In the case of a cylinder, which is the scenario presented in our exercise, we need to consider the shape's inherent symmetry and dimensions.

The height of the cylinder, given as 8 cm in our exercise, becomes the limit for the z-axis integration. The circular cross-section dictates the limits for the radial (r) and angular (θ) integrations - from 0 to the radius of the cylinder, and from 0 to 2π respectively. Integration is carried out in a step-wise manner, starting from the innermost integral which, for this cylinder, represents the radial component, and moving outward to the angular and vertical components.

By integrating over each dimension in a systematic fashion, we ensure the entire volume of the cylinder is considered, which is crucial for accurately calculating the mass of the water contained within.

Density Function

The density function characterizes how the density of a substance varies across its volume. In simple terms, it tells us how much mass is in a small piece of an object. A constant density is straightforward; for every tiny cube of material, you have the same amount of mass.

However, in this exercise, we are dealing with a variable density function, \(\rho(r)=1-0.05 e^{-0.01 r^{2}} \) grams per cubic centimeter. This means the density changes depending on the distance (r) from the central axis of the cylinder. Close to the axis, where the filament heats the water, the density is lower, represented by the exponential decay function. Understanding how to integrate this varying density over the volume of the cylinder is key to calculating the cylinder's mass accurately.

Integration by Parts

Integration by parts is a useful technique often employed when dealing with the product of two functions that are to be integrated. The technique is based on the product rule for differentiation and generally makes solving complex integrals more manageable.

In the context of the exercise, we applied integration by parts to solve the inner radial integration. With \( u=r \) and \( dv=(1 - 0.05 e^{-0.01r^2})dr \), the formula allowed us to break down the complex exponential function into simpler parts that could be evaluated. This technique transformed a challenging problem into smaller, more manageable steps, making it accessible for students to follow and understand.

Conceptually, integration by parts aligns closely with the idea of decomposition in mathematics; breaking down a complex concept into simpler ones to better tackle it.

Variable Density

Variable density refers to the characteristic of a material whose density is not uniform throughout its volume. This concept is vital in our cylinder mass calculation problem as the density changes as a function of distance from the cylinder's axis.

To manage this, the exercise demands integrating the density function across the variable 'r' within the limits of the cylinder's radius. The integration process thereby accounts for the change in density at each incremental radius, summing up the total mass contribution from all different radii.

Understanding variable density is critical in disciplines such as fluid dynamics, material science, and even astrophysics, where phenomena are rarely uniform in nature. In educational settings, such exercises help build students' ability to apply calculus to real-world, dynamically changing systems.