Problem 26 - Thomas Calculus

Question

Finding a centroid Find the centroid of the region between the \(x\) -axis and the arch \(y=\sin x, 0 \leq x \leq \pi\)

Step-by-Step Solution

Answer

The centroid is \(\left( \frac{\pi}{2}, \frac{\pi}{8} \right)\)."}

Step 1: Understand the Problem

To find the centroid of a region bounded by the curve \(y = \sin x\), the \(x\)-axis, from \(x = 0\) to \(x = \pi\), we need to calculate the coordinates \((\bar{x}, \bar{y})\) where \(\bar{x}\) is the average \(x\)-coordinate and \(\bar{y}\) is the average \(y\)-coordinate.

Step 2: Define the Centroid Formulas

The coordinates of the centroid \((\bar{x}, \bar{y})\) are given by:\[ \bar{x} = \frac{1}{A} \int_0^{\pi} x \sin x \, dx \]\[ \bar{y} = \frac{1}{A} \int_0^{\pi} \frac{1}{2} \sin^2 x \, dx \]where \(A\) is the area of the region. The area \(A\) is found by:\[ A = \int_0^{\pi} \sin x \, dx \]

Step 3: Calculate the Area Under the Curve

Calculate the area \(A\) of the region using the integral:\[ A = \int_0^{\pi} \sin x \, dx \]The antiderivative of \(\sin x\) is \(-\cos x\). Thus:\[ A = [-\cos x]_0^{\pi} = -\cos(\pi) + \cos(0) = 2 \]

Step 4: Calculate \(\bar{x}\)

Substitute into the formula for \(\bar{x}\):\[ \bar{x} = \frac{1}{2} \int_0^{\pi} x \sin x \, dx \]Integrate by parts, letting \(u = x\) and \(dv = \sin x \, dx\). You get \(du = dx\) and \(v = -\cos x\). Applying integration by parts:\[ \int x (-\cos x) \, dx = [-x \cos x]_0^{\pi} + \int \cos x \, dx \]\[ = (-\pi \cdot (-1) - 0 \cdot 1) + [-\sin x]_0^{\pi} \]\[ = \pi + 0 = \pi \]Substitute back:\[ \bar{x} = \frac{\pi}{2} \]

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

To find \(\bar{y}\):\[ \bar{y} = \frac{1}{2A} \int_0^{\pi} \sin^2 x \, dx \]Using the identity \(\sin^2 x = \frac{1 - \cos 2x}{2}\):\[ \bar{y} = \frac{1}{4} \int_0^{\pi} (1 - \cos 2x) \, dx \]\[ = \frac{1}{4} \left( [x]_0^{\pi} - \left[\frac{1}{2} \sin 2x\right]_0^{\pi} \right) \]\[ = \frac{1}{4} \left( \pi - 0 \right) = \frac{\pi}{4} \]Thus, \(\bar{y} = \frac{\pi}{4} \) averaged over \(2\) (factor in formula), so:\[ \bar{y} = \frac{\pi}{8} \]

Step 6: Conclusion

The coordinates of the centroid for the region between the \(x\)-axis and the curve \(y = \sin x\) from \(x = 0\) to \(x = \pi\) are \(\left( \frac{\pi}{2}, \frac{\pi}{8} \right)\).

Key Concepts

Integration by Parts

Integration by parts is a powerful technique for solving integrals, especially when the standard antiderivative is difficult to find. It's based on the product rule for differentiation and is especially handy when dealing with products of functions. Consider the integration by parts formula:\[\int u \, dv = uv - \int v \, du\]Here's how we practically use this formula:

• Choose a part of the integrand to be \(u\), which becomes simpler when differentiated.
• Assign what remains to be \(dv\), which is then integrated to find \(v\).
• Substitute \(u, v, du,\) and \(dv\) into the formula.

In our centroid problem, to find \(\bar{x}\), we let \(u = x\) (simplifies upon differentiation) and \(dv = \sin x \, dx\) (integrates easily to \(-\cos x\)). This strategic choice turns a complex integral into parts that are straightforward to handle. The integration by parts allows for breaking down integrals into simpler components that are easier to solve step-by-step.

Sinusoidal Functions

Sinusoidal functions, including sine and cosine, are fundamental in trigonometry. They describe waves and oscillations in mathematics and the natural world. The function \(y = \sin x\) represents a smooth, periodic wave that oscillates between -1 and 1, repeating every \(2\pi\). These properties make them useful in modeling phenomena like sound waves or alternating current.
An important feature of sinusoidal functions is their integration and differentiation properties. The derivative of \(\sin x\) is \(\cos x\), and the integral of \(\sin x\) is \(-\cos x\). This cyclical behavior is key when finding areas or centroids involving segments of sine curves.
In the context of our exercise, we consider the area under the curve of \(y = \sin x\) over the interval \(0 \leq x \leq \pi\). The area under this segment of the sine wave is symmetrical and represents half a wave. This helps in directly calculating the centroid by understanding sine's symmetry and periodic nature.

Area Under a Curve

The area under a curve is a foundational concept in calculus. It is used to find the size of the region between a curve and the x-axis, over a specified interval. The integral represents the accumulated area, providing insights into the size and relationships within mathematical functions.
For the function \(y = \sin x\) from \(x = 0\) to \(x = \pi\), the area under the curve encapsulates a crucial step in locating the centroid. This region forms the numerical 'weight' of the part of the graph under consideration. Specifically, we calculate the definite integral:\[A = \int_0^{\pi} \sin x \, dx\]Calculating this gives us an area of \(2\), essential for determining the average position \((\bar{x}, \bar{y})\) of the area—the centroid. This process of finding the area is linked to the concept of integration and provides a complete picture of the mathematical landscape involved.