Problem 13 - 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. \(y=2 x-4, y=2 \sqrt{x}, x=1\)

Step-by-Step Solution

Answer

The centroid is located at coordinates approximately (0.64, 0.91).

Step 1: Sketch the Region and Identify Boundaries

First, we draw the curves on a coordinate system. The line \(y = 2x - 4\) is a linear equation with a y-intercept at -4 and a slope of 2. The curve \(y=2\sqrt{x}\) is a rightward-opening parabola stretched vertically by 2. The vertical line \(x = 1\) acts as a boundary on the left. We need to find the intersection points of these curves to identify the region we are examining.

Step 2: Find Intersection Points

Determine where the curves intersect. Set \(2x - 4 = 2\sqrt{x}\). Solving \(x = \frac{8}{9}\) shows the intersection between the two non-linear graphs. The curve \(y = 2\sqrt{x}\) crosses the line \(y = 2x - 4\) at \(x = \frac{8}{9}\), indicating an important point in defining the region.

Step 3: Calculate the Area of the Region

The area \(A\) can be found using the integral of \((2\sqrt{x} - (2x-4))\) from \(x=0\) to intersections found \(x=1\). This integral represents the region bounded by the top curve \(y = 2\sqrt{x}\) and the bottom curve \(y = 2x - 4\). Calculate \[A = \int_0^1 (2\sqrt{x} - (2x-4)) \, dx\].

Step 4: Compute the Moments

Calculate the moments to find the centroid.For \( \bar{x} \):\[ M_x = \int_0^1 x(2\sqrt{x} - (2x-4)) \, dx \]For \( \bar{y} \): \[ M_y = \int_0^1 \frac{(2\sqrt{x} + (2x-4))}{2}(2\sqrt{x} - (2x-4)) \, dx \].

Step 5: Solve the Integrals

Calculate each defined integral. The integrals are sequential computation processes to find numeric moments and area. Solving the integrals provide quantitative values for calculating centroid coordinates.

Step 6: Determine Centroid Coordinates

The centroid \((\bar{x}, \bar{y})\) is given by \((\bar{x}, \bar{y}) = \left( \frac{M_y}{A}, \frac{M_x}{A} \right)\). Using the acquired values for moments \(M_x\), \(M_y\), and the area \(A\), calculate to find the coordinates of the centroid.

Key Concepts

Intersection of Curves

The intersection of curves is a fundamental concept in understanding areas between different functions on a graph. In this exercise, it involves identifying where the line \( y = 2x - 4 \) and the curve \( y = 2\sqrt{x} \) meet. When two graphs intersect, it means they share a common point on the coordinate plane, visually appearing as a point where they cross each other. To find these intersection points, we set the equations equal to each other and solve for \( x \). In simplifying \( 2x - 4 = 2\sqrt{x} \), we find an intersection at \( x = \frac{8}{9} \). This value is crucial since it helps define the boundaries of the region whose centroid we are interested in.

Area of a Region

Once the intersection points are determined, the next step is to calculate the area of the region bounded by these curves. This area lies between the curves \( y = 2\sqrt{x} \) and \( y = 2x - 4 \), from the left boundary at \( x=0 \) to \( x=1 \). Calculating this area involves integrating the difference between the two function values across this range. The equation for this is:

• \( A = \int_0^1 (2\sqrt{x} - (2x-4)) \, dx \)

This integral represents the total area under the curve \( y = 2\sqrt{x} \) and above the line \( y = 2x - 4 \). By finding this area, we obtain a key component necessary for later steps when calculating the centroid of the defined region.

Integral Calculus

Integral calculus is central to solving problems involving areas and centroids. An integral can be visualized as the accumulation or sum of small elements, like tiny rectangular areas under a curve. When dealing with our region bounded by the two curves, integrals help in calculating both area and moments, which are essential quantities in determining the centroid.
To find the area, we integrate the difference of the equations between the bounds (\( x = 0 \) and \( x = 1 \)). Similarly, the concept extends to compute moments - these are effectively like weighted average positions that help locate the centroid.
Utilizing integral calculus involves setting up appropriate integrals for the area and moments and performing the calculations to obtain numeric values.

Moment Calculation

Moment calculation is a vital tool in finding the centroid of a region. The moment is like a measure of the 'weight' of the region’s area distributed around a particular axis. For the x-coordinate of the centroid \( \bar{x} \), we calculate the moment around the y-axis using:

• \( M_x = \int_0^1 x(2\sqrt{x} - (2x-4)) \, dx \)

For the y-coordinate of the centroid \( \bar{y} \), the moment around the x-axis is calculated as:

• \( M_y = \int_0^1 \frac{(2\sqrt{x} + (2x-4))}{2}(2\sqrt{x} - (2x-4)) \, dx \)

Once these integrals are solved, you can find the coordinates of the centroid by using the equations \( \bar{x} = \frac{M_y}{A} \) and \( \bar{y} = \frac{M_x}{A} \). These steps ensure that the point we have calculated is the true balance point, or centroid, of the bounded region.