Problem 31 - Calculus Single Variable

Question

Find the center of \(\operatorname{mass}(\bar{x}, \bar{y})\) of the given region \(\mathcal{R},\) assuming that it has uniform unit mass density. \(\mathcal{R}\) is the region bounded above by \(y=2 \sqrt{x}\) and below by \(y=x\)

Step-by-Step Solution

Answer

The center of mass is \((-\frac{1}{20}, \frac{3}{2})\).

Step 1: Identify Boundaries

We start by identifying the bounding curves of the region \(\mathcal{R}\). The upper boundary is the curve \(y = 2 \sqrt{x}\) and the lower boundary is \(y = x\).

Step 2: Find Intersection Points

To find the limits of integration along the x-axis for our region, we set \(2 \sqrt{x} = x\) and solve for \(x\). Squaring both sides, we get \(4x = x^2\). Rearranging gives \(x^2 - 4x = 0\), or \(x(x - 4) = 0\). Thus, the intersection points are \(x = 0\) and \(x = 4\).

Step 3: Compute Area of Region \(\mathcal{R}\)

The area \(A\) of region \(\mathcal{R}\) is given by the integral of the difference between the upper and lower functions from \(x = 0\) to \(x = 4\). Thus, \(A = \int_0^4 (2\sqrt{x} - x) \, dx\).

Step 4: Evaluate the Area Integral

Compute \(\int_0^4 (2\sqrt{x} - x) \, dx\). Split it into two integrals: \(2\int_0^4 \sqrt{x} \, dx - \int_0^4 x \, dx\). These evaluate to \(2\left[ \frac{2}{3}x^{3/2} \right]_0^4 - \left[ \frac{x^2}{2} \right]_0^4\). Calculating, we find the area \(A = \frac{16}{3}\).

Step 5: Calculate \(\bar{x}\) - The x-coordinate of Center of Mass

Using the formula for \(\bar{x}\), we compute \(\bar{x} = \frac{1}{A} \int_0^4 x (2\sqrt{x} - x) \, dx\).

Step 6: Evaluate the Integral for \(\bar{x}\)

Compute \(\int_0^4 x (2\sqrt{x} - x) \, dx = \int_0^4 (2x\sqrt{x} - x^2) \, dx\). This splits into \(\int_0^4 2x^{3/2} \, dx - \int_0^4 x^2 \, dx\). Evaluating individually, we get \(\left[ \frac{4}{5}x^{5/2} \right]_0^4 - \left[ \frac{x^3}{3} \right]_0^4\). This gives \(\frac{32}{5} - \frac{64}{3} = \frac{32}{5} - \frac{192}{15} = \frac{-16}{15}\). Thus, \(\bar{x} = \frac{-16/15}{16/3} = \frac{-3}{60} = -\frac{1}{20}\).

Step 7: Calculate \(\bar{y}\) - The y-coordinate of Center of Mass

Use the formula for \(\bar{y}\), which is \(\bar{y} = \frac{1}{2A} \int_0^4 (2\sqrt{x} + x) (2\sqrt{x} - x) \, dx\).

Step 8: Evaluate the Integral for \(\bar{y}\)

Compute \(\int_0^4 (4x - x^2) \, dx\). Split into \(4\int_0^4 x \, dx - \int_0^4 x^2 \, dx\). The evaluation gives \(4\left[ \frac{x^2}{2} \right]_0^4 - \left[ \frac{x^3}{3} \right]_0^4\), which simplifies to \(32 - \frac{64}{3} = \frac{32}{3}\). Thus, \(\bar{y} = \frac{1}{2 \times 16/3} \cdot \frac{32}{3} = \frac{3}{2}\).

Step 9: Derive Center of Mass

Now that we have calculated both coordinates, the center of mass \((\bar{x}, \bar{y})\) is \((-\frac{1}{20}, \frac{3}{2})\).

Key Concepts

Mass Density

The concept of mass density is crucial when determining the center of mass of a region. In this exercise, we consider a region \(\mathcal{R}\) defined by uniform unit mass density. This means each point within the region has an equal mass contribution. To simplify calculations, assume mass density as 1 across the region:

• A uniform unit mass density uniformizes calculations, meaning each piece of the region contributes evenly to the overall mass.
• In cases of uniform density, the total mass is simply equivalent to the area of the region.

Understanding mass density allows us to focus on geometric properties, like area, instead of integrating an additional density term.

Intersection Points

Intersection points are vital in defining integration limits when calculating properties like area or center of mass.
For the given problem, we set the equations of the boundary curves equal to each other to find these points. The curves given are: \(y = 2\sqrt{x}\) (upper boundary) and \(y = x\) (lower boundary).

• Equate the two functions: \(2\sqrt{x} = x\).
• Square both sides: \(4x = x^2\).
• Rearrange to solve for \(x\): \(x^2 - 4x = 0\) or \(x(x - 4) = 0\).
• Intersection points are \(x = 0\) and \(x = 4\).

These points (0,0) and (4,4) give the domain of our integral calculations to capture the entire area of the region.

Integral Evaluation

Evaluating integrals helps us find the area and center of mass of a region. For the center of mass, we compute two main integral types:

• To find area: the integral of the difference between bounding curves.
• To find \(\bar{x}\) and \(\bar{y}\): weighted integrals that involve the variable being evaluated.

Steps include:

• Calculate area: \A = \int_0^4 (2\sqrt{x} - x) \, dx\ leading to **Area, A = \frac{16}{3}**.
• Calculate \(\bar{x}\): \bar{x} = \frac{1}{A} \int_0^4 x (2\sqrt{x} - x) \, dx\, obtaining **\bar{x} = -\frac{1}{20}**.
• Calculate \(\bar{y}\): \bar{y} = \frac{1}{2A} \int_0^4 (x)(2\sqrt{x} - x) \, dx\ leading to **\bar{y} = \frac{3}{2}**.

Each step simplifies a broader calculation into manageable integrals that isolate x and y components of the center of mass.

Bounding Curves

Bounding curves define the perimeter of the region of interest by determining where the mass density is concentrated. For this exercise, these curves are crucial:

• The upper boundary is \(y = 2\sqrt{x}\).
• The lower boundary is \(y = x\).

These curves enclose the particular section of the plane by setting the vertical limits:

• The area under the \(y = 2\sqrt{x}\) curve minus the area under \(y = x\) gives the desired region.
• Integrals defined between these curves quantify properties such as area and center of mass.

By focusing on the difference between these bounding curves over the x-domain, we effectively capture the full area and mass distribution of the region.