Problem 30
Question
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. $$g(x)=0, \quad f(x)=2+\sin x, \quad x=0, \quad \text { and } \quad x=2 \pi$$ $$\text { (Hint: } \int x \sin x d x=\sin x-x \cos x+C$$
Step-by-Step Solution
Verified Answer
The centroid is \((\pi, 2)\).
1Step 1: Understand the Problem
We need to find the centroid of a region bounded by the curves \(g(x) = 0\) and \(f(x) = 2 + \sin x\) between \(x = 0\) and \(x = 2\pi\). The centroid's coordinates \((\bar{x}, \bar{y})\) can be found using formulas for the centroid: \(\bar{x} = \frac{1}{M} \int_{a}^{b} x [f(x) - g(x)] \, dx\) and \(\bar{y} = \frac{1}{M} \int_{a}^{b} \frac{1}{2} [f(x)^2 - g(x)^2] \, dx\), where \(M\) is the area of the region.
2Step 2: Calculate the Area (M)
The area \(M\) under the curve \(f(x) = 2 + \sin x\) from \(x = 0\) to \(x = 2\pi\) is given by \(M = \int_{0}^{2\pi} (2 + \sin x) \, dx\). Use the integral: \(\int \sin x \, dx = -\cos x + C\) to find\[ M = \int_{0}^{2\pi} 2 \, dx + \int_{0}^{2\pi} \sin x \, dx = \left[ 2x \right]_{0}^{2\pi} + \left[ -\cos x \right]_{0}^{2\pi} = 4\pi + 0 = 4\pi.\]
3Step 3: Calculate \(\bar{x}\)
\(\bar{x}\) is given by\[ \bar{x} = \frac{1}{4\pi} \int_{0}^{2\pi} x(2 + \sin x) \, dx = \frac{1}{4\pi} \left[ 2\int_{0}^{2\pi} x \, dx + \int_{0}^{2\pi} x \sin x \, dx \right]. \] From the hint, \(\int x \sin x \, dx = \sin x - x \cos x + C\), find\[ = \left[ 2 \frac{x^2}{2} \right]_{0}^{2\pi} + \left[ x \sin x + \cos x \right]_{0}^{2\pi} = \left[ x^2 \right]_{0}^{2\pi} + 0 = \frac{(2\pi)^2}{4\pi} = \pi.\]
4Step 4: Calculate \(\bar{y}\)
\(\bar{y}\) is given by\[ \bar{y} = \frac{1}{4\pi} \int_{0}^{2\pi} \frac{1}{2}((2 + \sin x)^2 - 0^2) \, dx = \frac{1}{4\pi} \int_{0}^{2\pi} \left( 4 + 4\sin x + \sin^2 x \right) \, dx. \] Using \(\sin^2 x = \frac{1 - \cos 2x}{2}\), find\[ = \frac{1}{4\pi} \int_{0}^{2\pi} \left( 4 + 4\sin x + \frac{1 - \cos 2x}{2} \right) \, dx = \frac{1}{4\pi} \left[ 8x - 2 \right]_{0}^{2\pi} = 2. \]
5Step 5: Combine Results
With values \(\bar{x} = \pi\) and \(\bar{y} = 2\), the centroid of the given region is \((\pi, 2)\).
Key Concepts
Definite IntegralArea under a CurveTrigonometric FunctionsCoordinate Geometry
Definite Integral
The definite integral helps us calculate quantities that are accumulated over continuous ranges, such as areas under curves. The process involves determining the integral of a function between two specified limits, known as the bounds of integration. This allows us to find the "net" area, taking into account any parts of the curve that dip below the x-axis.
To use definite integrals for calculating areas, consider an integral \(\int_{a}^{b} f(x) \, dx\), where \(f(x)\) is your function and \(a\) and \(b\) are the limits. This represents the area between the curve and the x-axis from \(a\) to \(b\). In our centroid calculation, we first find the total area \(M\) by integrating the function \(2 + \sin x\) from \(0\) to \(2\pi\), resulting in \(M = 4\pi\).
With practice, definite integrals become a powerful tool for solving problems in calculus. By focusing on preparation and process, they can transform intricate calculations into manageable solutions.
To use definite integrals for calculating areas, consider an integral \(\int_{a}^{b} f(x) \, dx\), where \(f(x)\) is your function and \(a\) and \(b\) are the limits. This represents the area between the curve and the x-axis from \(a\) to \(b\). In our centroid calculation, we first find the total area \(M\) by integrating the function \(2 + \sin x\) from \(0\) to \(2\pi\), resulting in \(M = 4\pi\).
With practice, definite integrals become a powerful tool for solving problems in calculus. By focusing on preparation and process, they can transform intricate calculations into manageable solutions.
Area under a Curve
Finding the area under a curve is a common application of integration in calculus, which helps in various fields ranging from physics to economics. This involves using integrals to determine how much space is beneath the curve of a function across a certain interval.
In the exercise provided, finding the area under the function \(f(x)=2 + \sin x\) involves evaluating the integral \(\int_{0}^{2\pi} (2 + \sin x) \, dx\). We split this into two parts for convenience: \(\int_{0}^{2\pi} 2 \, dx\) and \(\int_{0}^{2\pi} \sin x \, dx\). The evaluation results in an area \(M = 4\pi\), which represents the total area between the curve and the x-axis within the specified limits.
Understanding areas in this way provides insights into both geometric and practical applications, helping students comprehend physical spaces or economic quantities modeled by such functions.
In the exercise provided, finding the area under the function \(f(x)=2 + \sin x\) involves evaluating the integral \(\int_{0}^{2\pi} (2 + \sin x) \, dx\). We split this into two parts for convenience: \(\int_{0}^{2\pi} 2 \, dx\) and \(\int_{0}^{2\pi} \sin x \, dx\). The evaluation results in an area \(M = 4\pi\), which represents the total area between the curve and the x-axis within the specified limits.
Understanding areas in this way provides insights into both geometric and practical applications, helping students comprehend physical spaces or economic quantities modeled by such functions.
Trigonometric Functions
Trigonometric functions like sine and cosine are fundamental in calculus, particularly when dealing with periodic phenomena. They describe wave-like behaviors and are integral in calculations involving rotations and oscillations.
In the given problem, we deal with the function \(f(x) = 2 + \sin x\), which incorporates the sine function. The sine function, \(\sin x\), fluctuates between -1 and 1 with a period of \(2\pi\), adding a dynamic aspect to the linear part (which is simply \(2\)). This combination results in oscillations of the entire function, shifting it upwards by 2 units.
When solving integrals that contain trigonometric elements, familiarity with identities and properties can simplify calculations. For example, the hint provided uses the identity of the integral of \(x \sin x\), aiding in the derivation for calculating the centroid effectively. A solid grasp of trigonometric functions ensures that evaluating these integrals does not become overly complex.
In the given problem, we deal with the function \(f(x) = 2 + \sin x\), which incorporates the sine function. The sine function, \(\sin x\), fluctuates between -1 and 1 with a period of \(2\pi\), adding a dynamic aspect to the linear part (which is simply \(2\)). This combination results in oscillations of the entire function, shifting it upwards by 2 units.
When solving integrals that contain trigonometric elements, familiarity with identities and properties can simplify calculations. For example, the hint provided uses the identity of the integral of \(x \sin x\), aiding in the derivation for calculating the centroid effectively. A solid grasp of trigonometric functions ensures that evaluating these integrals does not become overly complex.
Coordinate Geometry
Coordinate geometry, or analytic geometry, allows us to precisely locate points and shapes in a plane using algebraic equations. It is essential for analyzing the properties of figures defined by algebraic constraints, as seen in centroid calculations.
The concept of a centroid refers to the geometric center of a shape—in our exercise, it's the region bounded by the functions \(g(x) = 0\) and \(f(x) = 2 + \sin x\) from \(x = 0\) to \(x = 2\pi\). We leverage the formulas for the centroid's x-coordinate (\(\bar{x}\)) and y-coordinate (\(\bar{y}\)) which involve definite integrals to locate this balance point.
For \(\bar{x}\), we compute \(\int x (2 + \sin x) \, dx\) to find the average distance along the x-axis. Similarly, for \(\bar{y}\), we use \(\int \frac{1}{2}((2 + \sin x)^2) \, dx\) to determine the height of the centroid. The intersection of these results gives the centroid's position, revealing the balance of the shape. Mastery of coordinate geometry provides the framework for exploring such interactions between algebra and geometry.
The concept of a centroid refers to the geometric center of a shape—in our exercise, it's the region bounded by the functions \(g(x) = 0\) and \(f(x) = 2 + \sin x\) from \(x = 0\) to \(x = 2\pi\). We leverage the formulas for the centroid's x-coordinate (\(\bar{x}\)) and y-coordinate (\(\bar{y}\)) which involve definite integrals to locate this balance point.
For \(\bar{x}\), we compute \(\int x (2 + \sin x) \, dx\) to find the average distance along the x-axis. Similarly, for \(\bar{y}\), we use \(\int \frac{1}{2}((2 + \sin x)^2) \, dx\) to determine the height of the centroid. The intersection of these results gives the centroid's position, revealing the balance of the shape. Mastery of coordinate geometry provides the framework for exploring such interactions between algebra and geometry.
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