Question

Find the centroid of the region bounded by the given curves. Make a sketch and use symmetry where possible. $$ y=x^{3}, y=0, x=1 $$

Short answer

The centroid is at \(\left( \frac{4}{5}, \frac{2}{7} \right)\).

Step-by-step solution

Understand the Problem

We need to find the centroid of the region bounded by the curves \( y = x^3 \), \( y = 0 \), and \( x = 1 \). The centroid is the point \((\bar{x}, \bar{y})\) that represents the center of mass of the region.

Visualize the Region

First, sketch the region based on the given curves. The curve \( y = x^3 \) starts at the origin (0,0) and increases steeply as \( x \) increases to 1. The line \( y = 0 \) is the x-axis, and \( x = 1 \) is a vertical line stopping the region at \( x = 1 \). The region is a section of the first quadrant bounded by these curves from \( x = 0 \) to \( x = 1 \).

Set Up Centroid Formulas

The formulas for the centroid \( (\bar{x}, \bar{y}) \) of a region bounded by \( y = f(x) \), \( y = 0 \), from \( x = a \) to \( x = b \) are:\[ \bar{x} = \frac{1}{A} \int_{a}^{b} x f(x) \, dx \]\[ \bar{y} = \frac{1}{2A} \int_{a}^{b} (f(x))^2 \, dx \]where \( A \) is the area of the region.

Compute the Area A

Calculate the area \( A \) under the curve \( y = x^3 \) from \( x = 0 \) to \( x = 1 \):\[ A = \int_{0}^{1} x^3 \, dx = \left. \frac{x^4}{4} \right|_0^1 = \frac{1^4}{4} - \frac{0^4}{4} = \frac{1}{4} \]

Compute \( \bar{x} \)

Substitute the known values into the \( \bar{x} \) formula:\[ \bar{x} = \frac{1}{A} \int_{0}^{1} x(x^3) \, dx = 4 \int_{0}^{1} x^4 \, dx \]\[ = 4 \left. \frac{x^5}{5} \right|_0^1 = 4 \left( \frac{1}{5} \right) = \frac{4}{5} \]

Compute \( \bar{y} \)

Substitute the known values into the \( \bar{y} \) formula:\[ \bar{y} = \frac{1}{2A} \int_{0}^{1} (x^3)^2 \, dx = \frac{1}{2 \times \frac{1}{4}} \int_{0}^{1} x^6 \, dx \]\[ = 2 \left. \frac{x^7}{7} \right|_0^1 = 2 \left( \frac{1}{7} \right) = \frac{2}{7} \]

Result Interpretation

The calculated centroid \((\bar{x}, \bar{y}) = \left( \frac{4}{5}, \frac{2}{7} \right)\) is the balance point of the region bounded by the curves. This makes sense given the shape's asymmetrical nature with respect to the x-axis.

Key concepts

Calculus and Centroids

Calculus is a branch of mathematics that deals with the study of change and motion. It involves concepts like derivatives, which measure rates of change, and integrals, which measure accumulation.
In the context of finding the centroid of a region, calculus helps us solve problems involving shapes and figures that are not necessarily regular. The centroid of a region is essentially its center of mass or balance point. To locate this point, we use integration, a fundamental tool of calculus.
By integrating the function that defines the region, we can determine the area under the curve and compute the coordinates of the centroid, (\(\bar{x}\),\(\bar{y}\)). Understanding and applying calculus principles is crucial for solving problems related to centroids, especially when the boundaries are defined by complex curves.

The Role of Integration

Integration is the mathematical process of finding the whole from its parts. It's a crucial concept in calculus that we use to measure areas, volumes, and other quantities. Integration is particularly useful when dealing with irregular shapes, such as the region under the curve\(y = x^3\).
To find the centroid of such a region, integrating the function over its limits gives the total area. In this exercise, the integral (\(\int_{0}^{1} x^3 \, dx\))calculates the area under the curve from \(x = 0\) to \(x = 1\). With the area known, integration helps us determine \(\bar{x}\) and \(\bar{y}\), using the centroid formulas:

• \(\bar{x} = \frac{1}{A} \int_{a}^{b} x f(x) \, dx\)
• \(\bar{y} = \frac{1}{2A} \int_{a}^{b} (f(x))^2 \, dx\)

This application of integration is essential for analyzing and understanding the spatial properties of various regions, especially when bounded by nonlinear curves.

Understanding Area Under a Curve

The area under a curve represents the 'accumulated' geometric measure between the curve and the x-axis over a specified interval. In this exercise, we are interested in the area under the curve \(y = x^3\) from \(x = 0\) to \(x = 1\).
Using integration, the area is calculated as\(A = \int_{0}^{1} x^3 \, dx = \frac{1}{4}\).This value represents the space that the curved line and the x-axis enclose. Knowing the area is crucial for determining the centroid, as it helps find the average distance along the x and y axes.
When we say 'area under a curve,' we often consider the physical or visual aspect of the space on graphs, which gives insight into the region's size. It is a foundational concept in calculus and a necessary step in locating centroids for various shapes.

Exploring the First Quadrant

The first quadrant of the Cartesian plane is where both x and y are positive. This quadrant is important as it often involves problems that deal with real-world applications where negative values don't make sense, e.g., distances or areas.
In the context of this exercise, the region under examination lies entirely in the first quadrant. The curve \(y = x^3\), along with the x-axis and the line \(x = 1\), defines the area in this quadrant.
Mathematically, when working in the first quadrant, we're often interested in the properties of shapes and figures bounded by curves that don't extend into negative x or y values. This focus simplifies calculations for centroids, helping to avoid complications introduced by negative numbers and maintain consistent positive spatial relationships in these problems.