Problem 24 - Thomas Calculus

Question

Finding a centroid The area of the region in the first quadrant bounded by the parabola \(y=6 x-x^{2}\) and the line \(y=x\) is 125\(/ 6\) square units. Find the centroid.

Step-by-Step Solution

Answer

The centroid is at \((3, \frac{12}{5})\)."

Step 1: Find Intersections

To find the intersections of the curve and the line, we equate the given equations: \(6x - x^2 = x\). Simplify this to \(-x^2 + 5x = 0\), and factor it to \(x(-x + 5) = 0\). The solutions are \(x = 0\) and \(x = 5\). Therefore, the region is bounded between \(x = 0\) and \(x = 5\).

Step 2: Set Up Integrals for the Centroid

The formulas for the centroid \((\bar{x}, \bar{y})\) of a region in the plane are \(\bar{x} = \frac{1}{A} \int_a^b x(f(x) - g(x)) \ dx\) and \(\bar{y} = \frac{1}{A} \int_a^b \frac{1}{2}((f(x))^2 - (g(x))^2) \ dx\). Here, \(f(x) = 6x - x^2\) and \(g(x) = x\), and the area \(A = \frac{125}{6}\).

Step 3: Calculate \(\bar{x}\)

Plug the values into the \(\bar{x}\) formula: \(\bar{x} = \frac{6}{125} \int_0^5 x((6x - x^2) - x) \ dx = \frac{6}{125} \int_0^5 x(5x - x^2) \ dx = \frac{6}{125} \int_0^5 (5x^2 - x^3) \ dx\). Integrate to get: \(\bar{x} = \frac{6}{125} \left[\frac{5}{3} x^3 - \frac{1}{4} x^4 \right]_0^5 = \frac{6}{125}(\frac{5}{3} \, 125 - \frac{1}{4} \, 625) = 3\).

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

Use the formula for \(\bar{y}\): \(\bar{y} = \frac{6}{125} \int_0^5 \frac{1}{2}((6x-x^2)^2 - x^2) \ dx\). First, compute \((6x-x^2)^2 = 36x^2 - 12x^3 + x^4\). Hence, \(\bar{y} = \frac{3}{125} \int_0^5 (36x^2 - 12x^3 + x^4 - x^2) \ dx = \frac{3}{125} \int_0^5 (35x^2 - 12x^3 + x^4) \ dx\). Integrate and evaluate: \(\bar{y} = \frac{3}{125} \left[\frac{35}{3} x^3 - 3 x^4 + \frac{1}{5} x^5 \right]_0^5 = \frac{3}{125}(\frac{35}{3} * 125 - 3 * 625 + 25) = \frac{12}{5}\).

Step 5: Conclusion

The centroid \((\bar{x}, \bar{y})\) of the region is calculated to be \(\left(3, \frac{12}{5}\right)\).

Key Concepts

Integral Calculus

Integral calculus is a branch of calculus that deals with the accumulation of quantities and finding areas under curves. It is a way to calculate the total size or value, given a continuous rate of change. For instance, when you find the integral of a function over an interval, you are essentially finding the area under the curve of that function between two points. This is central to understanding physical quantities like distance, area, and volume.

In our exercise of finding the centroid, integral calculus helps us determine the location of the centroid by evaluating the entire shape formed by the intersections of the curves. The integrals are used to bind the differences in the functional curves between their intersection points.

Centroid Formulas

Centroid formulas allow us to find the point that represents the center of mass or geometric center of a region. For any bounded region in the plane, the centroid's coordinates \(\bar{x}, \bar{y}\) can be calculated using the formulas:

• \(\bar{x} = \frac{1}{A} \int_a^b x(f(x) - g(x)) \, dx\)
• \(\bar{y} = \frac{1}{A} \int_a^b \frac{1}{2}((f(x))^2 - (g(x))^2) \, dx\)

Here, \(f(x)\) and \(g(x)\) are the functions defining the upper and lower boundaries of our region, "a" and "b" are the points that bound the region along the x-axis, and "A" is the total calculated area of the region. By integrating these expressions, the centroid formulas provide the average position of all points in the shape. These calculations are essential as they assist in determining the point that will balance the shape, meaning every part of the shape has equally distributed mass or area around the centroid.

Parabola and Line Intersection

To find where a parabola intersects with a line, we equate their equations. This is like finding common solutions to the two equations, which gives the points where one curve meets the other. The intersection points are key in understanding the region's boundaries. In our problem, the parabolic function is \(y = 6x - x^2\) and the line is \(y = x\).

By setting these equations equal, \(6x - x^2 = x\), we can simplify and solve to find \(x = 0\) and \(x = 5\). These intersections are where the parabola and the line cross, forming a closed region on the x-axis that defines the limits for the integral calculations. Solving these intersections correctly establishes the area over which you will compute the centroid and can often be the first critical step in problems like these.

Area under a Curve

Calculating the area under a curve involves integration between two limits. This process allows us to determine the size or extent of a region bounded by a curve. In our case, the area under the curve is essential for further calculations, such as finding the centroid.

Between the points of intersection, we calculate the area by integrating the difference between the upper function \(f(x)\) and the lower function \(g(x)\) as follows: \(A = \int_a^b (f(x) - g(x)) \, dx\). In this problem, \(f(x)\) is the parabola \(y=6x-x^2\) and \(g(x)\) is the line \(y=x\). Calculating this helps in finding the total region under the composite shape formed by the parabola and line. Finding the area is a prerequisite for calculating the centroid since it normalizes the x and y positional calculations.