Problem 38 - Thomas Calculus

Question

In Exercises \(35-38,\) find the centroid of the thin plate bounded by the graphs of the given functions. Use Equations (6) and (7) with \(\delta=1\) and \(M=\) area of the region covered by the plate. $$\begin{array}{l}{g(x)=0, \quad f(x)=2+\sin x, \quad x=0, \quad \text { and } \quad x=2 \pi} \\ {\text { (Hint: } \int x \sin x d x=\sin x-x \cos x+C . )}\end{array}$$

Step-by-Step Solution

Answer

Centroid: \((\pi/2, 1)\)."

Step 1: Set Up Variables for Functions and Limits

We are given the functions \( g(x) = 0 \) and \( f(x) = 2 + \sin x \) with limits \( x = 0 \) and \( x = 2\pi \). We start by defining these results explicitly as part of the domain for calculating the area.

Step 2: Calculate the Area Between the Curves

The area \( M \) is obtained by integrating the difference between the two functions from \( x = 0 \) to \( x = 2\pi \). This gives us:\[M = \int_{0}^{2\pi} [f(x) - g(x)]\, dx = \int_{0}^{2\pi} (2 + \sin x) \, dx.\] Evaluate the integral: \[M = \left[ 2x - \cos x \right]_{0}^{2\pi} = 2(2\pi) - (\cos(2\pi) - \cos(0)) = 4\pi.\]

Step 3: Calculate the Centroid's x-coordinate

For the x-coordinate, \( \overline{x} \), use the formula:\[\overline{x} = \frac{1}{M} \int_{0}^{2\pi} x(f(x) - g(x)) \, dx = \frac{1}{4\pi} \int_{0}^{2\pi} x(2 + \sin x) \, dx.\]Split the integrals:\[= \frac{1}{4\pi} \left(\int_{0}^{2\pi} 2x \, dx + \int_{0}^{2\pi} x \sin x \, dx\right).\]For the integral of \( 2x \), solve:\[\frac{1}{4\pi} \left[ x^2 \right]_{0}^{2\pi} = \frac{1}{4\pi}((2\pi)^2 - 0) = \pi.\]For the integral of \( x \sin x \), note the hint:\[\int x \sin x \, dx = \sin x - x \cos x + C\]Evaluate:\[\frac{1}{4\pi} \left[ -x \cos x + \sin x \right]_{0}^{2\pi} = 0.\]Thus, \(\overline{x} = \frac{1}{4\pi}(\pi + 0) = \frac{1}{4\pi}(2\pi^2) = \pi/2.\)

Step 4: Calculate the Centroid's y-coordinate

For the y-coordinate, \( \overline{y} \), use the formula:\[\overline{y} = \frac{1}{2M} \int_{0}^{2\pi} (f(x)^2 - g(x)^2) \, dx = \frac{1}{8\pi} \int_{0}^{2\pi} (2 + \sin x)^2 \, dx.\]Expand and solve the integral:\[\int_{0}^{2\pi} (4 + 4\sin x + \sin^2 x) \, dx = \int_{0}^{2\pi} 4\, dx + \int_{0}^{2\pi} 4\sin x \, dx + \int_{0}^{2\pi} \frac{1 - \cos(2x)}{2} \, dx.\]Evaluate each part:1. \( \int_{0}^{2\pi} 4 \, dx = 8\pi \).2. \( \int_{0}^{2\pi} 4\sin x \, dx = 0 \).3. \( \int_{0}^{2\pi} \frac{1}{2} \, dx - \frac{1}{2}\int_{0}^{2\pi} \cos(2x) \, dx = 0\).Combine to find \( \overline{y} \):\[\frac{1}{8\pi}(8\pi) = 1.\]

Step 5: Provide the Final Answer

The centroid of the thin plate bounded by the given functions is at the point \((\pi/2, 1)\).

Key Concepts

Integration

Integration is a fundamental mathematical concept used to find areas under curves or the total accumulation of quantities. In this exercise, integration helps determine both the area between the curves and centroid coordinates. The process involves summing small slices, under the curve, to approximate the whole.
For instance, to find the area between two curves, you integrate the difference of their functions over a set interval. Here, we used the integration \[ \int_{0}^{2\pi} (2 + \sin x) \, dx \] in calculating the area of the region. This integral adds up all slice heights of the curve \( f(x) = 2 + \sin x \) over the interval from \( x = 0 \) to \( x = 2\pi \).
Integration plays a crucial role in determining centroid coordinates. To find \( \overline{x} \) and \( \overline{y} \), you'll integrate expressions involving \( x \) and \( y \) weighted by respective areas. So, integration is a compass guiding us to these calculations.

Area Between Curves

Understanding the area between curves is pivotal in finding centroids, which requires precise calculations. Consider two functions, \( f(x) \) and \( g(x) \). The area between these curves, over an interval \([a, b]\), is determined by integrating the difference: \[ \int_{a}^{b} [f(x) - g(x)] \, dx. \]
In our example, \( g(x) = 0 \) simplifies calculations since we're effectively integrating the single function \( f(x) \).This means adding up all values that the function \( f(x) = 2 + \sin x \) takes from \( x = 0 \) to \( x = 2 \pi \).The area (\( 4\pi \)) serves not only for calculating the region size but also in the computations for the coordinates of the centroid.
By understanding the area between curves, we utilize geometry to pave the way for deeper applications like centroid determination.

Trigonometric Integrals

Trigonometric integrals appear frequently when dealing with functions involving sine or cosine. Solving these integrals requires applying specific techniques, often using identities or substitution.
For instance, the integral \( \int x \sin x \, dx \) was solved using a known result: \( \sin x - x \cos x + C \).Solutions for such integrals are essential as they facilitate area or centroid calculations involving trigonometric functions.
Solving trigonometric integrals requires a solid grasp of trigonometric identities and integral rules. They might involve breaking down the integrals into simpler parts or employing mathematical tricks like using anti-derivative tables. This knowledge ultimately aids in revealing geometric properties such as periodic trends or symmetry.

Centroid Coordinates

The centroid of a region is its geometric center, akin to finding its balance point. Determining centroid coordinates involves a methodical approach using both integration and the area calculated in previous steps.
For the x-coordinate, \( \overline{x} \), we apply \[ \overline{x} = \frac{1}{M} \int_{0}^{2\pi} x(f(x) - g(x)) \, dx, \]where \( M \) is the area previously calculated. By breaking it into manageable parts, the solutions guide us to find the \( \overline{x} = \pi/2. \)
Similarly, the y-coordinate, \( \overline{y} \), derives from \[ \overline{y} = \frac{1}{2M} \int_{0}^{2\pi} (f(x)^2 - g(x)^2) \, dx. \]It's finding the average height over the interval, allowing us to calculate \( \overline{y} = 1. \)
So, centroids are located using an integration-assisted approach, which not only uncovers the coordinates \((\pi/2, 1)\) but provides insights into the overall balance of the shape.