Problem 12 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the centroid of the region bounded by the given curves. Make a sketch and use symmetry where possible. $$ y=\frac{1}{2}\left(x^{2}-10\right), y=0, \text { and between } x=-2 \text { and } x=2 $$

Step-by-Step Solution

Answer

The centroid of the region is at \( (0, 6.75) \).

Step 1: Sketch the Region and Identify Symmetry

First, consider the equation \( y = \frac{1}{2}(x^2 - 10) \) and \( y = 0 \). The parabolic equation describes a downward-opening parabola, and \( y=0 \) is the x-axis. Sketch these on a Cartesian plane focusing between \( x = -2 \) and \( x = 2 \). Note that the region is symmetric about the y-axis because \( y \) is an even function.

Step 2: Determine Intersection Points

Calculate the points of intersection of \( y = \frac{1}{2}(x^2 - 10) \) with \( y = 0 \). Setting these equal, solve \( \frac{1}{2}(x^2 - 10) = 0 \), which simplifies to \( x^2 - 10 = 0 \). Solving \( x^2 = 10 \) leads to \( x = \pm \sqrt{10} \). However, we are considering \( x = -2 \) to \( x = 2 \), so we do not reach these points.

Step 3: Find the Area of the Region

The area \( A \) of the region is found by integrating \( y = \frac{1}{2}(x^2 - 10) \) from \( x=-2 \) to \( x=2 \):\[ A = \int_{-2}^{2} \frac{1}{2}(x^2 - 10) \, dx \]Calculate\[ A = \frac{1}{2} \int_{-2}^{2} (x^2 - 10) \, dx = \frac{1}{2} \left[ \frac{x^3}{3} - 10x \right]_{-2}^{2} \]This evaluates to \( \frac{1}{2} \left[ (\frac{8}{3} - 20) - (-\frac{8}{3} + 20) \right] \), resulting in \(-\frac{32}{3}\). Take the absolute value since area is positive: \( \frac{32}{3} \).

Step 4: Calculate the Y-coordinate of the Centroid

The y-coordinate \( \bar{y} \) of the centroid is given by:\[ \bar{y} = \frac{1}{A} \int_{-2}^{2} \frac{1}{2}(x^2 - 10)^2 \, dx \]First, find the constant factor: \( \frac{1}{\left| A \right|} = \frac{3}{32} \), since \( A \) is \( \frac{32}{3} \). Now integrate and evaluate:\[ \int_{-2}^{2} \frac{1}{2}(x^2 - 10)^2 \, dx = \int_{-2}^{2} \left( \frac{1}{4}(x^2 - 10)^2 \right) \, dx \].Computing this integral gives a result that simplifies to 72. Thus, \( \bar{y} = \frac{3}{32} \times 72 = 6.75 \).

Step 5: Conclusion by Symmetry for X-coordinate

Due to symmetry of the region about the y-axis, the region is symmetrical, implying the x-coordinate of the centroid \( \bar{x} = 0 \). Hence, the coordinates of the centroid are \( (0, 6.75) \).

Key Concepts

Symmetry in Calculus

Symmetry is a powerful concept in calculus that can simplify finding centroids and integrating functions. When a shape or a function is symmetrical, it can reduce the complexity of computations. In our problem, the region bounded by the curve \( y = \frac{1}{2}(x^2 - 10) \) and the line \( y = 0 \) exhibits symmetry about the y-axis. This is because the function \( y = \frac{1}{2}(x^2 - 10) \) is an even function, meaning it is identical on both sides of the axis.

• If a region is symmetrical about the y-axis, the x-coordinate of its centroid is \( \bar{x} = 0 \), as confirmed by the symmetry of its graph.
• This symmetry allows us to only calculate the y-coordinate, making the procedure efficient.
• Recognizing such patterns can help you avoid unnecessary calculations, speeding up problem-solving processes.

Symmetry in calculus not only saves time but also gives insights into the inherent properties of functions and shapes. Next time you approach a similar problem, look for symmetry—it might be the key to a simpler solution.

Integration in Calculus

Integration is a fundamental concept in calculus, crucial for finding areas under curves, and is indispensable in locating centroids. In the given exercise, we integrate to find the area and y-coordinate of the centroid of the region bounded by the given curves.

• The integration of \( y = \frac{1}{2}(x^2 - 10) \) between \( x = -2 \) and \( x = 2 \) provides the area of the region. This involves calculating the definite integral:

\[A = \int_{-2}^{2} \frac{1}{2}(x^2 - 10) \, dx\] Evaluating this integral involves finding antiderivatives and subsequently applying the limits \(-2\) and \(2\). This process yields the total area \( A \), essential for further calculations.

• Incorrect integration boundaries or miscalculations can lead to negative areas, but absolute values rectify these errors, ensuring positive areas.
• Similarly, we use integration to find the y-coordinate of the centroid \( \bar{y} \), assessing how far away the center of mass is vertically from the x-axis.

Mastering integration techniques will enhance your ability to solve diverse calculus problems, and verify solutions with contextual sense.

Centroid in Mathematics

The centroid is a crucial concept in mathematics, particularly in geometry and calculus. It represents the point where the region's mass is evenly balanced. Finding the centroid is akin to finding the average position of all the points in a plane region, considering it as a physical body.

• This concept is widely applied, from designing structures to calculating centers of mass in physics.
• In our problem, the x-coordinate of the centroid was zero due to symmetry, simplifying calculations as only the y-coordinate needed detailed work.

For finding the y-coordinate \( \bar{y} \) of a centroid, we compute:\[\bar{y} = \frac{1}{A} \int_{-2}^{2} \frac{1}{2}(x^2 - 10)^2 \, dx\] This expression weighs the function's squared value (representing height and density) over the defined range.

• The result \( (0, 6.75) \) signifies that the centroid lies on the y-axis. Here, x reveals a direct relationship between area density and balance position.

Understanding centroids helps grasp the equilibrium of distributed systems, fundamental in comprehending both real-world physical and mathematical concepts.