Problem 41 - 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 \(y=\left(1-x^{2}\right)^{3 / 2}\) and below by the \(x\) -axis.

Step-by-Step Solution

Answer

The center of mass is at (0, 0.3).

Step 1: Identify the Region of Integration

The region \( \mathcal{R} \) is defined by the function \( y = (1 - x^2)^{3/2} \) and is above the x-axis. To find where this region exists, identify the interval for \( x \). This function is real and non-negative while \( 1-x^2 \) is non-negative, which means the region is bounded between \( x = -1 \) and \( x = 1 \).

Step 2: Determine the Area of the Region

The area of the region, \( A \), is needed to find the center of mass. Compute \( A \) using the integral: \[ A = \int_{-1}^{1} (1-x^2)^{3/2} \, dx \] To evaluate this integral, make the substitution \( x = \sin \theta \), which transforms the limits from \( x = -1 \) to \( x = 1 \) into \( \theta = -\frac{\pi}{2} \) to \( \theta = \frac{\pi}{2} \). After substitution and solving, this integral evaluates to \( \frac{8}{3} \). Thus, the area \( A = \frac{8}{3} \).

Step 3: Calculate the Moments and Coordinates for the Center of Mass

For a region with uniform density, the center of mass coordinates can be calculated using the formulas: \[ \bar{x} = \frac{1}{A} \int_{-1}^{1} x(1-x^2)^{3/2} \, dx \] \[ \bar{y} = \frac{1}{2A} \int_{-1}^{1} (1-x^2)^3 \, dx \] Calculate \( \bar{x} \): The integral evaluates to zero since \( x(1-x^2)^{3/2} \) is an odd function and the limits of integration are symmetric around zero. Thus, \( \bar{x} = 0 \).Calculate \( \bar{y} \): The integral simplifies by using another trigonometric substitution or direct integration. The computation yields \( \bar{y} = \frac{3}{10} \).

Step 4: Conclusion

By combining the results, the center of mass of the region \( \mathcal{R} \) is \( (\bar{x}, \bar{y}) = (0, \frac{3}{10}) \).

Key Concepts

Integral Calculus in Finding the Area and Mass

Integral calculus is a powerful mathematical tool used to compute areas, volumes, and related measures. In this exercise, we use it to find the area of a region under the curve defined by the function \[ y = (1-x^2)^{3/2} \] above the x-axis. This is crucial for determining the center of mass of the region.
To begin, we identify the interval for integration by examining where the function is real and non-negative, which occurs between \( x = -1 \) and \( x = 1 \). The area \( A \) is then calculated using the integral:
\[ A = \int_{-1}^{1} (1-x^2)^{3/2} \, dx \] Such integrals often involve non-standard functions and require special techniques, making them common integral calculus challenges.
Through trigonometric substitution, where \( x = \sin \theta \), we can simplify this integral, translating it to a form that is easier to solve. The limits of integration change from \( x = -1 \) to \( x = 1 \) to \( \theta = -\frac{\pi}{2} \) to \( \theta = \frac{\pi}{2} \). After substitution and evaluation, we find the area to be \( \frac{8}{3} \), a crucial value for later steps in finding the center of mass.

Understanding Moment of Inertia and Center of Mass

The concept of moment of inertia is typically associated with rotational motion, but it also plays a vital role in calculating the center of mass of a region. In this scenario, we aim to find the coordinates \( \bar{x} \) and \( \bar{y} \) of the center of mass.
The moment about the y-axis helps to find \( \bar{x} \) using:

• \( \bar{x} = \frac{1}{A} \int_{-1}^{1} x(1-x^2)^{3/2} \, dx \)

This integral evaluates to zero due to symmetry and the odd nature of the function, meaning our x-coordinate for the center of mass is 0.
For \( \bar{y} \), we compute the integral:

• \( \bar{y} = \frac{1}{2A} \int_{-1}^{1} (1-x^2)^3 \, dx \)

This process may involve another level of trigonometric substitution or direct integration strategies to simplify. Ultimately, this yields \( \bar{y} = \frac{3}{10} \). By combining these computations, we determine the region's center of mass at \( (0, \frac{3}{10}) \). Understanding these moments thus allows us to pinpoint where the center of mass lies.

Applying Trigonometric Substitution in Integration

Trigonometric substitution is a key technique in integral calculus used to simplify integrals involving roots and quadratics, as seen in the exercise.When dealing with integrals like:

• \( \int (1-x^2)^{3/2} \, dx \)

Oftentimes, substituting \( x = \sin \theta \) helps by transforming the expression under the square root into a form involving trigonometric identities \( 1 - \sin^2 \theta = \cos^2 \theta \).
This substitution not only simplifies the integral but also changes the variable of integration from \( x \) to \( \theta \). In this specific exercise, it translates the limits to \( \theta = -\pi/2 \) to \( \pi/2 \).Through carefully evaluating such substitutions and working through the resulting trigonometric identities, we convert more complex integrals into manageable forms. This method is also non-trivial but provides clear pathways for evaluating the definite integrals required to solve our problem. Thus, trigonometric substitution serves as a bridge between intricate algebraic functions and simpler trigonometric ones in many real-world applications.