Problem 19 - 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 mass density. \(\mathcal{R}\) is the first quadrant region bounded above by \(y=\sqrt{4-x^{2}}\) and below by the \(x\) -axis.

Step-by-Step Solution

Answer

The center of mass is at \((0, \frac{4}{3\pi})\).

Step 1: Identify the Region

The region \( \mathcal{R} \) is in the first quadrant and is bounded above by the curve \( y = \sqrt{4-x^2} \) and below by the \( x \)-axis. This describes the upper half of a circle with radius 2 centered at the origin.

Step 2: Use Symmetry of the Region

Since the region \( \mathcal{R} \) is symmetric with respect to the \( y \)-axis, the center of mass will lie on the \( y \)-axis. This implies that \( \bar{x} = 0 \).

Step 3: Determine the Limits for Integration

The region \( \mathcal{R} \) extends along the \( x \)-axis from \( x=0 \) to \( x=2 \). These are the limits for the integration used to find \( \bar{y} \).

Step 4: Find the Total Mass of the Region

Because \( \mathcal{R} \) has uniform density, we can treat its mass as proportional to its area. The area is equal to the integral \( \int_0^2 \sqrt{4-x^2} \, dx \).

Step 5: Calculate \( \bar{y} \)

\( \bar{y} \) is given by \( \frac{1}{A} \int_0^2 y^2 \, dy \), with \( y = \sqrt{4-x^2} \). Therefore, \( \bar{y} = \frac{1}{A} \int_0^2 \sqrt{4-x^2} \cdot x \, dx \).

Step 6: Evaluate Integrals

To compute the integrals, make a trigonometric substitution: let \( x = 2\sin(\theta) \), yielding \( dx = 2\cos(\theta) \, d\theta \). Consequently, \( \sqrt{4 - x^2} = 2\cos(\theta) \). The integral \( \int_0^2 \sqrt{4-x^2} \, dx\) becomes \( \int_0^{\frac{\pi}{2}} 4\cos^2(\theta) \, d\theta \). It simplifies to \( 2\theta + \sin(2\theta) \) evaluated from \( 0 \) to \( \frac{\pi}{2} \).

Step 7: Final Calculations and Results

The complete calculation is \( A = \int_0^2 \sqrt{4-x^2} \, dx = \frac{\pi}{2} \). Therefore, \( \bar{y} = \frac{1}{ \frac{\pi}{2} } \cdot \frac{4}{3} = \frac{4}{3\pi} \). The coordinates of the center of mass are \( (\bar{x}, \bar{y}) = (0, \frac{4}{3\pi}) \).

Key Concepts

Uniform Mass Density

In physics and mathematics, when we mention uniform mass density, we are assuming that the mass is evenly distributed throughout the region. This means every unit of area has the same mass. In our example region \( \mathcal{R} \), being in the first quadrant of the circle, this concept simplifies calculations because we don't need to consider varying mass. The centroid of such a uniform distribution often aligns with geometric centers like the midpoint or center of a circle.

• Uniform mass density allows us to relate mass to area directly.
• This significantly reduces the need for complex, variable-density calculations.

Given that \( \mathcal{R} \) has uniform density, the total mass of the region can be directly calculated using its area. This results in straightforward integration to find the center of mass.

Integration Limits

Integration limits define the bounds within which we perform our integrations. It is a critical step in solving problems involving calculus, as it frames the region over which calculations are made. In our problem:

• The region \( \mathcal{R} \) is bounded by \( x = 0 \) and \( x = 2 \), so these become the limits of integration.
• These boundaries ensure that all calculations perfectly capture the precise area under consideration.

This means our integrals for calculating area or mass only consider parts of the region that contribute to the final result. Without correctly setting these limits, our integration could provide inaccurate values that don't apply directly to the problem.

Trigonometric Substitution

Trigonometric substitution is a technique often used in calculus to simplify integrals that involve square roots of quadratic expressions. In our problem, we use this method to simplify the expression \( \sqrt{4 - x^2} \).

• We substitute \( x = 2\sin(\theta) \) to help simplify the integral \( \int_0^2 \sqrt{4-x^2} \, dx \).
• This results in \( dx = 2\cos(\theta) \, d\theta \), which transforms our integral into a more manageable form.

With these transformations, the integral becomes \( \int_0^{\frac{\pi}{2}} 4\cos^2(\theta) \, d\theta \). Such substitutions are particularly useful as they convert complicated algebraic expressions into simpler trigonometric functions, making integration straightforward.

Symmetric Region

Symmetry in mathematics can greatly simplify calculations. When we refer to a symmetric region, as with \( \mathcal{R} \), we mean that the area, mass, or shape is mirrored or balanced in some way. For our problem:

• The region \( \mathcal{R} \) is symmetric about the \( y \)-axis.
• This symmetry implies that the center of mass lies directly on this axis, simplifying our calculation for \( \bar{x} \) to be 0.

Using symmetry saves effort since it identifies natural balancing points in the region. This simplification is particularly useful as it reduces computational complexity, offering quick insights into properties like the center of mass when other characteristics are uniformly distributed.