Question

As needed, use a computer to plot graphs and to check values of integrals. Find the centroid of the first quadrant part of the arc \(x^{2 / 3}+y^{2 / 3}=a^{2 / 3} .\) Hint: Let \(x=a \cos ^{3} \theta, y=a \sin ^{3} \theta\).

Short answer

The centroid is at \(\left(\frac{4a}{15}, \frac{4a}{15}\right)\).

Step-by-step solution

- Parametrizing the arc

Use the hint given to parametrize the arc. Let \(x = a \cos^3(\theta)\) and \(y = a \sin^3(\theta)\). Recall that the parameter \(\theta\) ranges from \(0\) to \(\frac{\pi}{2}\) as the arc is in the first quadrant.

- Express the differential elements

Express the differentials \(dx\) and \(dy\): For \(x\), \(dx = a \times 3 \cos^2(\theta) (-\sin(\theta)) d\theta = -3a \cos^2(\theta) \sin(\theta) d\theta\). For \(y\), \(dy = a \times 3 \sin^2(\theta) \cos(\theta) d\theta = 3a \sin^2(\theta) \cos(\theta) d\theta\).

- Set up integral for arc length

The differential length \(ds\) is given by \(ds = \sqrt{\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2} d\theta\). Calculate accordingly: \( \frac{dx}{d\theta} = -3a \cos^2(\theta) \sin(\theta)\) and \( \frac{dy}{d\theta} = 3a \sin^2(\theta) \cos(\theta) \). Now, \(ds = \sqrt{(-3a \cos^2(\theta) \sin(\theta))^2 + (3a \sin^2(\theta) \cos(\theta))^2} d\theta\ = \sqrt{9 a^2 \cos^4(\theta) \sin^2(\theta) + 9 a^2 \sin^4(\theta) \cos^2(\theta)} d\theta\ = 3a \cos(\theta) \sin(\theta) d\theta\).

- Integrating to find the centroid

The centroid \(( \bar{x}, \bar{y} )\) coordinates are given by \( \bar{x} = \frac{1}{L} \int_{C} x ds \) and \( \bar{y} = \frac{1}{L} \int_{C} y ds \). Here \(L\) is the arc length found by integrating \(ds\) from \(0\) to \(\frac{\pi}{2}\): \( L = \int_0^{\pi/2} 3a \cos(\theta) \sin(\theta) d\theta\ = \frac{3a}{2}\).

- Finding \( \bar{x} \)

Integrate to find \( \bar{x} \): \( \bar{x} = \frac{1}{L} \int_{0}^{\pi/2} a \cos^3(\theta) \cdot 3a \cos(\theta) \sin(\theta) d\theta \ = \frac{2}{a} \int_{0}^{\pi/2} 3a^2 \cos^4(\theta) \sin(\theta) d\theta = \frac{2}{a} \cdot \frac{3a^2}{5}\ = \frac{6a}{5}\ \times \frac{2}{3} = \frac{4a}{15}\).

- Finding \( \bar{y} \)

Integrate to find \( \bar{y} \): \( \bar{y} = \frac{1}{L} \int_{0}^{\pi/2} a \sin^3(\theta) \cdot 3a \sin(\theta) \cos(\theta) d\theta \ = \frac{2}{a} \int_{0}^{\pi/2} 3a^2 \sin^4(\theta) \cos(\theta) d\theta \ = \frac{2}{a} \cdot \frac{3a^2}{5}\ = \frac{6a}{5}\ \times \frac{2}{3} = \frac{4a}{15}\).

Key concepts

Arc Length Calculation Using Integration

Finding the arc length of a curve involves an integral calculation. To start, we often use parametrized equations for simplicity. For an arc defined parametrically, such as \( x = a \cos^3(\theta) \) and \( y = a \sin^3(\theta) \), we need to express differentials \( dx \) and \( dy \).
This formulates into \( ds \), the differential length along the curve, expressed as \( ds = \sqrt{( \frac{dx}{d\theta} )^2 + ( \frac{dy}{d\theta} )^2} d\theta \).
Imagine it like finding the hypotenuse in tiny triangles along the curve.
Next, we plug in our arc's differential elements to integrate and find the total arc length from \( \theta = 0 \) to \( \theta = \frac{\pi}{2} \). This integration gives the full arc length needed for further geometry calculations, like finding centroids.

Understanding Parametric Equations

Parametric equations describe a set of coordinates in terms of one or more variables called parameters. Here, we utilized \( x = a \cos^3(\theta) \) and \( y = a \sin^3(\theta) \), where \( \theta \) is our parameter varying from 0 to \( \frac{\pi}{2} \).
This method is particularly useful in dealing with more complex curves that are difficult to define solely with \( x \) and \( y \).
Parametric forms make it easier to compute various properties like arc lengths or centroids by manipulating simpler trigonometric functions.
Make sure to correctly define the range of the parameter to properly describe the entire arc or curve.

Integral Applications in Centroid Calculation

Calculating centroids involves applying integrals to weighted average positions. The centroid of a curve, or arc, can be found through specific integral formulas.
Given the arc length \( L \), the coordinates of the centroid \( (\bar{x}, \bar{y}) \) are derived from

• \( \bar{x} = \frac{1}{L} \int_{C} x ds \)
• \( \bar{y} = \frac{1}{L} \int_{C} y ds \)

In practice, parametric equations represent \( x \) and \( y \), integrated with respect to \( \theta \).
Each part of the curve must be integrated considering the differential arc length component, \( ds \), to ensure accurate centroid placement that balances all parts of the curve. Integrals in these formulations average out the positional distances, giving the centroid's exact coordinates.