Calculating The Mass Of An Oversized Hockey Puck With Variable Density
Hey guys! Ever wondered how to calculate the mass of something when its density isn't uniform? Let's dive into a cool problem involving an oversized hockey puck where the density changes depending on the distance from the center. We'll use some calculus magic to figure this out. Buckle up!
Understanding the Problem
So, we've got this hockey puck, right? But it's not your regular, evenly-dense puck. This one's got a density that varies. Imagine it's made of some funky material where the stuff closer to the edge is either heavier or lighter than the stuff in the middle. We're given that the puck has a radius of 2 inches, which is a key piece of information. The density, denoted by the Greek letter rho (ρ), is described by the function ρ(x) = x³ - 3x + 7. This function tells us the density at any point in the puck, where 'x' represents the distance from that point to the very center (the origin) of the puck. This is super important because it tells us that the density isn't constant; it changes as we move away from the center. The goal here is to find the total mass of this puck. Now, why can't we just multiply the density by the area? Because the density isn't the same everywhere! That's where integration comes in handy. Integration allows us to add up infinitely small pieces of the puck, each with its own density, to get the total mass. Think of it like adding up the weights of a bunch of tiny rings, each with a slightly different density, to get the total weight of the puck. Understanding this varying density is crucial because it dictates how we approach the mass calculation. We can't just use a simple formula; we need to use calculus to account for this change. This problem is a classic example of how calculus can be used to solve real-world physics problems, and it's a great way to flex your math muscles!
Setting Up the Integral
To calculate the mass, we're going to use a technique that involves slicing the puck into infinitesimally thin rings. Think of it like cutting a pizza into infinitely many concentric rings. Each ring has a tiny width, which we'll call dr, and a radius r, which is the distance from the center. The area of each ring is approximately the circumference (2πr) multiplied by the width (dr), giving us dA = 2πr dr. This is the key to setting up our integral. We're essentially breaking down the puck into tiny pieces whose areas we can easily calculate. Now, we know the density ρ(x) at any distance x from the origin, and in this case, x is the same as our radius r. So, the density at a ring of radius r is ρ(r) = r³ - 3r + 7. To find the mass of each tiny ring (dm), we multiply the density by the area: dm = ρ(r) dA = (r³ - 3r + 7)(2πr dr). This gives us an expression for the mass of an infinitesimally small ring at a distance r from the center. To find the total mass (M) of the entire puck, we need to add up the masses of all these tiny rings. This is where integration comes in. We integrate the expression for dm over the radius of the puck, which ranges from 0 (the center) to 2 inches (the edge). So, our integral looks like this: M = ∫dm = ∫(r³ - 3r + 7)(2πr dr), where the integral is taken from 0 to 2. This integral represents the sum of the masses of all the infinitesimal rings that make up the puck. Setting up the integral correctly is crucial because it translates the physical problem into a mathematical one that we can solve. Once we evaluate this integral, we'll have the total mass of our oversized hockey puck!
Evaluating the Integral
Alright, let's get our hands dirty and evaluate the integral we set up in the last section. We have M = ∫₀² (r³ - 3r + 7)(2πr dr). First, let's pull the constant 2π out of the integral: M = 2π ∫₀² (r³ - 3r + 7)r dr. Now, we need to distribute the r inside the integral: M = 2π ∫₀² (r⁴ - 3r² + 7r) dr. This simplifies our integrand and makes it easier to integrate term by term. Next, we find the antiderivative of each term: The antiderivative of r⁴ is r⁵/5, the antiderivative of -3r² is -r³, and the antiderivative of 7r is (7/2)r². So, we have M = 2π [r⁵/5 - r³ + (7/2)r²]₀². Now, we need to evaluate this expression at the upper limit (2) and the lower limit (0) and subtract the results. At r = 2, we have (2⁵/5) - 2³ + (7/2)(2²) = 32/5 - 8 + 14 = 32/5 + 6 = 32/5 + 30/5 = 62/5. At r = 0, all the terms are zero, so the expression evaluates to 0. Therefore, M = 2π (62/5 - 0) = 2π (62/5) = 124π/5. This gives us the exact value of the mass. To get a decimal approximation, we can plug this into a calculator: M ≈ 124π/5 ≈ 77.91 inches. So, the mass of our oversized hockey puck is approximately 77.91 units of mass (since we weren't given specific units for density, we'll just leave it as units of mass). This calculation demonstrates the power of calculus in solving problems involving variable densities and complex shapes. We successfully used integration to sum up the masses of infinitely many tiny rings to find the total mass of the puck. Pretty cool, right?
Interpreting the Result
Okay, so we've calculated the mass to be approximately 77.91 units. But what does this number actually tell us? Well, first of all, it's important to remember that we didn't have specific units for the density function ρ(x), so our mass is in generic