Problem 50 - Calculus Single Variable

Question

Find the center of mass of the given region \(\mathcal{R},\) assuming that it has uniform unit mass density. \(\mathcal{R}\) is the region bounded above by $$ y=\left\\{\begin{array}{cl} x^{2} & \text { if } 0 \leq x \leq 2 \\ 4(x-3)^{2} & \text { if } 2 \leq x \leq 3 \end{array}\right. $$ and below by the \(x\) -axis.

Step-by-Step Solution

Answer

The center of mass is at \((1.875, 0.75)\).

Step 1: Determine boundaries of each section

The region \( \mathcal{R} \) is divided into two parts: \( 0 \leq x \leq 2 \) where the boundary line is \( y = x^2 \), and \( 2 \leq x \leq 3 \) where the boundary is \( y = 4(x-3)^2 \). Identify these boundaries for further calculations.

Step 2: Calculate area of each section

The area \( A_1 \) for \( 0 \leq x \leq 2 \) can be determined by integrating \( y = x^2 \) from 0 to 2: \[ A_1 = \int_0^2 x^2 \, dx = \frac{x^3}{3} \Big|_0^2 = \frac{8}{3} \]For the portion \( 2 \leq x \leq 3 \), the area \( A_2 \) can be computed as: \[ A_2 = \int_2^3 4(x-3)^2 \, dx = \frac{4}{3}(x-3)^3 \Big|_2^3 = \frac{4}{3} \].

Step 3: Compute total area of the region

Combine the areas of both sections to find the total area of \( \mathcal{R} \):\[ A = A_1 + A_2 = \frac{8}{3} + \frac{4}{3} = 4 \].

Step 4: Find the centroid of each section

For the \( x \) from 0 to 2, find the \( x \) and \( y \) centroids:\[ \bar{x}_1 = \frac{1}{A_1} \int_0^2 x \cdot x^2 \, dx = \frac{3}{8} \int_0^2 x^3 \, dx = \frac{3}{8} \cdot \frac{16}{4} = \frac{3}{2} \]\[ \bar{y}_1 = \frac{1}{A_1} \int_0^2 \frac{x^2}{2} \cdot x^2 \, dx = \frac{3}{8} \int_0^2 x^4 \, dx = \frac{9}{10} \]For \( x \) from 2 to 3:\[ \bar{x}_2 = \frac{1}{A_2} \int_2^3 x \cdot 4(x-3)^2 \, dx = 2.5 \]\[ \bar{y}_2 = \frac{1}{A_2} \int_2^3 \frac{4(x-3)^2}{2} \cdot 4(x-3)^2 \, dx = \frac{4}{5} \].

Step 5: Calculate the coordinates of the center of mass

Combine the centroids to determine the overall center of mass of \( \mathcal{R} \) as follows:\[ \bar{x} = \frac{1}{A}(A_1\bar{x}_1 + A_2\bar{x}_2) = \frac{1}{4} \left(\frac{8}{3} \cdot \frac{3}{2} + \frac{4}{3} \cdot 2.5 \right) = 1.875 \]\[ \bar{y} = \frac{1}{A}(A_1\bar{y}_1 + A_2\bar{y}_2) = \frac{1}{4} \left(\frac{8}{3} \cdot \frac{9}{10} + \frac{4}{3} \cdot \frac{4}{5} \right) = 0.75 \].

Key Concepts

Centroid Calculation

The centroid of a region is often referred to as the 'center of mass' when the density is uniform. In mathematical terms, this point is like the average position of all the points of a shape. For 2D shapes, centroids are needed in problems involving geometrical properties of a region. We find the centroid by finding the mean of the x-coordinates and the y-coordinates, which are determined separately for each defined region. To find the x-coordinate of the centroid, the formula used is:

• \[ \bar{x} = \frac{1}{A} \int x \, dA \]

For the y-coordinate, it's:

• \[ \bar{y} = \frac{1}{A} \int y \, dA \]

In our exercise, the centroid of each section of the region \( \mathcal{R} \) is calculated by applying these formulas over the specified intervals created by the piecewise functions.When the entire figure contains multiple sections like in this exercise, we determine the centroid of each subsection separately. Then, we compute an overall centroid considering the total area.The overall centroid's coordinates will hence have the contributions of the centroids of individual sections weighted by their respective areas.

Area Under Curves

Finding the area under a curve is a central activity when calculating centroids, since it directly ties into mass properties like the center of mass. We use integration techniques to find these areas, as integration effectively sums up infinitesimally small areas beneath a curve to compute a total area.For any given function, the area under its curve from a point \( a \) to \( b \) on the x-axis is calculated using:

• \[ \text{Area} = \int_{a}^{b} f(x) \, dx \]

In our specific problem, this is done twice because the region \( \mathcal{R} \) is bounded by two different curves on different intervals. Calculating these areas correctly is crucial since they contribute to calculating the center of mass by establishing each segment's weight in the area.

Integration Techniques

Integration is the mathematical technique used to sum an infinite series of small elements to find the total. Different functions require different integration methods, and knowing which method to apply is essential. Here, we primarily use basic techniques such as the power rule for integration.The integration process essentially does the following:

• For a function \( x^n \), integrate it as: \[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \]

In our region \( \mathcal{R} \), calculating the areas involved integrating polynomial functions. These integrations involved power rules, shifting functions, and knowing where to apply definite limits.Understanding integration techniques is vital to navigating the calculations needed to calculate centroids and other geometric properties.

Piecewise Functions

Piecewise functions are functions that have different definitions over distinct intervals. They are essential in modeling situations where a function’s rule changes based on the input.In this exercise, the region \( \mathcal{R} \) is described using a piecewise function:

• From \( 0 \leq x \leq 2 \), the boundary is given by \( y = x^2 \).
• From \( 2 \leq x \leq 3 \), the boundary changes to \( y = 4(x-3)^2 \).

When working with piecewise functions, it's important to correctly identify the different domains and apply the appropriate function accordingly. Piecewise functions require careful treatment since the integrals or calculations must be computed separately for each piece. Each segment's centroids and areas are computed individually before finally combining them for the complete region.