Problem 35 - 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. $$g(x)=x^{2} \quad \text { and } \quad f(x)=x+6$$

Step-by-Step Solution

Answer

The centroid is at \(\left(\frac{17}{5}, \frac{31}{5}\right)\)."

Step 1: Understand the Problem

We are tasked with finding the centroid of a region bounded by two functions, \(g(x) = x^2\) and \(f(x) = x + 6\). A centroid is the geometric center of a region.

Step 2: Find Points of Intersection

To find the points of intersection between the curves \(g(x) = x^2\) and \(f(x) = x + 6\), we set them equal to each other: \(x^2 = x + 6\). Solving this equation using algebraic techniques, we get the points of intersection as \(x = -2\) and \(x = 3\).

Step 3: Calculate the Area of the Region

The area \(M\) of the region bounded by the curves from \(x = -2\) to \(x = 3\) is calculated as \(M = \int_{-2}^{3} [(x + 6) - x^2] \, dx\). Solving this integral, we find \(M = \frac{125}{6}\).

Step 4: Compute the Centroid Coordinates – \(\bar{x}\)

The \(\bar{x}\) coordinate of the centroid is given by the formula \(\bar{x} = \frac{1}{M} \int_{-2}^{3} x[(x + 6) - x^2] \, dx\). Solving this integral gives \(\bar{x} = \frac{17}{5}\).

Step 5: Compute the Centroid Coordinates – \(\bar{y}\)

The \(\bar{y}\) coordinate of the centroid is given by \(\bar{y} = \frac{1}{M} \int_{-2}^{3} \frac{1}{2}[(x + 6)^2 - (x^2)^2] \, dx\). Calculating this, we get \(\bar{y} = \frac{31}{5}\).

Step 6: Conclusion

Thus, the centroid of the region bounded by the curves \(g(x) = x^2\) and \(f(x) = x + 6\) is \(\left(\frac{17}{5}, \frac{31}{5}\right)\).

Key Concepts

Area Under Curve

The concept of the area under a curve is fundamental in integral calculus. Imagine you are looking at a graph with a curve plotted on it. The region underneath this curve, stretching over a certain interval on the x-axis, is what we aim to calculate. It’s like finding the total amount of space or coverage the curve takes up on the graph.

To calculate this area, we use integrals. For instance, to find the area under the curve between two points (say from \(x = a\) to \(x = b\)), we calculate the definite integral of the function with respect to x over the interval \[ \text{Area} = \int_{a}^{b} f(x) \, dx. \] This idea is very similar to counting how many unit squares fit under the curve, except we use calculus to do it smoothly and accurately.

In our problem, this technique helped find the area between two curves: \(x^2\) and \(x + 6\), simplifying it by finding where they intersect and calculating the definite integral between these points.

Integral Calculus

Integral calculus is all about adding things up. It deals with integrals, which are essentially a way to sum continuously distributed quantities. Think of it as finding the total of tiny infinite parts that make up a whole body, like calculating the whole pizza by considering each tiny slice.

An integral solution helps us find quantities such as areas under curves, volumes, central points like the centroid, and much more. In our exercise, integrals were used to determine both the total area under the curves and the centroid coordinates.

For finding the centroid, integrals helped calculate the weighted average of the region. The process needs \[ \bar{x} = \frac{1}{M} \int_{a}^{b} x[f(x) - g(x)] \, dx \] for the x-coordinate and for the y-coordinate \( \bar{y} = \frac{1}{2M} \int_{a}^{b} [(f(x))^2 - (g(x))^2] \, dx \). This involves breaking the region into infinitesimally small parts and calculating their contributions accurately.

Coordinate Geometry

Coordinate geometry is the study of geometry using a coordinate system. It combines algebra and geometry to understand spatial structures and relations. This is the foundation for precisely determining where points, lines, and regions lie in space based on numbers and equations.

In our context, we use coordinate geometry to accurately describe the curves given by \( g(x) = x^2 \) and \( f(x) = x + 6 \). By finding where they intersect on the graph, we define the boundaries of the region we're examining.

This also sets the stage for integrating within those limits. Essentially, it allows us to transform geometric properties into algebraic equations, making it easier to perform calculations such as finding centroids or areas. This approach gives clear spatial insight through the lens of numbers.