Problem 9 - Calculus: Early Transcendentals

Question

A thin plate lies in the region contained by \(y=x^{1 / 3}\) and the \(x\) -axis between \(x=0\) and \(x=1 .\) Find the centroid.

Step-by-Step Solution

Answer

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

Step 1: Understand the Problem

We need to find the centroid of a region bounded by the curve \( y = x^{1/3} \) and the \( x \)-axis from \( x = 0 \) to \( x = 1 \). A centroid is the geometric center of a region.

Step 2: Calculate Area of the Region

The area \( A \) is given by the integral: \[ A = \int_{0}^{1} x^{1/3} \ dx. \] We need to compute this integral to find the area of the region.

Step 3: Evaluate the Integral for Area

Calculate the integral: \[ A = \int_{0}^{1} x^{1/3} \ dx. \] This evaluates to \( \frac{3}{4} \).

Step 4: Determine the \( x \)-coordinate of the Centroid

The \( x \)-coordinate of the centroid \( \bar{x} \) is given by:\[ \bar{x} = \frac{1}{A} \int_{0}^{1} x \cdot x^{1/3} \ dx = \frac{1}{A} \int_{0}^{1} x^{4/3} \ dx. \] Substitute \( A = \frac{3}{4} \) to find \( \bar{x} \).

Step 5: Evaluate the \( x \)-coordinate Integral

Compute the integral: \[ \bar{x} = \frac{4}{3} \int_{0}^{1} x^{4/3} \ dx. \] This evaluates to \( \frac{4}{3} \times \frac{3}{7} = \frac{4}{7} \).

Step 6: Determine the \( y \)-coordinate of the Centroid

The \( y \)-coordinate of the centroid \( \bar{y} \) is given by:\[ \bar{y} = \frac{1}{A} \int_{0}^{1} \frac{1}{2} (x^{1/3})^2 \ dx \] \[ = \frac{1}{A} \int_{0}^{1} \frac{1}{2} x^{2/3} \ dx. \] Use the expression for \( A = \frac{3}{4} \).

Step 7: Evaluate the \( y \)-coordinate Integral

Compute the integral: \[ \bar{y} = \frac{4}{3} \int_{0}^{1} \frac{1}{2} x^{2/3} \ dx. \] This evaluates to \( \frac{4}{3} \times \frac{3}{5} \times \frac{1}{2} = \frac{2}{5} \).

Step 8: Conclusion

The centroid of the region bounded by \( y = x^{1/3} \) and the \( x \)-axis between \( x = 0 \) and \( x = 1 \) is \( \left( \frac{4}{7}, \frac{2}{5} \right) \).

Key Concepts

Integral Calculation

Understanding integrals is key when dealing with calculus problems like finding the centroid of a region. An integral helps us find the accumulated total of a function over a given interval. In this particular problem, we need to evaluate an integral to find the area under a curve, which is crucial for obtaining the centroid of the specified region.

To determine the area, we used the integral \( A = \int_{0}^{1} x^{1/3} \, dx \). This integral calculates the area beneath the curve \( y = x^{1/3} \). Solving this integral provides the area of the entire region of interest. Through this calculation, the integral evaluates to \( \frac{3}{4} \).

Integrals break down the computation into infinitely small parts and sum them up, yielding the total area or other values of interest. This step is foundational not only for finding areas but also for solving many other calculus problems.

Area under a Curve

The area under a curve represents the total space between the curve and the axis over a specific range. This concept is fundamental when we look at determining the characteristics of a region, like finding its centroid. Understanding this area helps us find geometric or physical properties of the shape formed by the curve and the axes.

In our example, the area is bounded by the curve \(y = x^{1/3}\) and the \(x\)-axis, between \(x = 0\) and \(x = 1\). To compute this, we evaluated \( \int_{0}^{1} x^{1/3} \, dx = \frac{3}{4} \), which gives us the area under the curve. This step is essential, as knowing the area is a prerequisite for calculating the centroid coordinates of the region.

This understanding extends beyond just curves—it applies to any situation where you need to find how much a function "covers" over an interval.

Geometric Center

The geometric center, or centroid, of a region is like its "balance point." It's where the region would balance perfectly if it were made of a uniform material. For planar regions bound by curves, it's necessary to find both \(x\) and \(y\) coordinates to specify this center.

To find the centroid of the region defined by the curve \( y = x^{1/3} \) and the \(x\)-axis, we compute two integrals: one for the \(x\)-coordinate and one for the \(y\)-coordinate.

**1. \(x\)-coordinate \( \bar{x} \):**
We use \( \bar{x} = \frac{1}{A} \int_{0}^{1} x \cdot x^{1/3} \, dx \). Substituting in the area \( A = \frac{3}{4} \), we find the \(x\)-coordinate as: \( \frac{4}{7} \).

**2. \(y\)-coordinate \( \bar{y} \):**
This is found using \( \bar{y} = \frac{4}{3} \int_{0}^{1} \frac{1}{2} x^{2/3} \, dx \), resulting in \( \frac{2}{5} \).

Combining these, the centroid is at \(( \frac{4}{7}, \frac{2}{5} )\), giving a precise location of the geometric center of the area defined by the curve and axes.