Problem 22 - Single Variable Calculus

Question

Find the centroid of the region bounded by the graphs of the given equations. $$ y=x^{3}, \quad y=\sqrt[3]{x}, \quad x=0, \quad x=1 $$

Step-by-Step Solution

Answer

The centroid of the region bounded by the given equations is \(\left(\frac{61}{56}, \frac{16}{35}\right)\).

Step 1: Identify bounding curves and set up integrals

First, let's identify the curves bounding the region: - Left boundary: \(x=0\) - Right boundary: \(x=1\) - Lower curve: \(y=x^3\) - Upper curve: \(y=\sqrt[3]{x}\) Now we will set up the integrals for the x-coordinate of the centroid, \(\bar{x}\), and the y-coordinate of the centroid, \(\bar{y}\): \[- A=\int_a^b (f(x) - g(x)) dx \] \[ \bar{x} =\frac{1}{A}\int_a^b x(f(x) - g(x)) dx \] \[ \bar{y} =\frac{1}{A}\int_a^b \frac{1}{2}(f^2(x) - g^2(x)) dx \] Where: - \(f(x)\) = upper curve, in this case, \(\sqrt[3]{x}\) - \(g(x)\) = lower curve, in this case, \(x^3\) - \(a\) = left boundary, in this case, \(0\) - \(b\) = right boundary, in this case, \(1\)

Step 2: Find the area A

Now let's calculate the area A using the formula: \[ A = \int_0^1 (\sqrt[3]{x} - x^3) dx \] We can compute the integral: \[A = \left[\frac{3}{4}x^{4/3} - \frac{1}{4}x^4 \right]_0^1 =\left(\frac{3}{4}(1)^{4/3} - \frac{1}{4}(1)^4\right)-\left(\frac{3}{4}(0)-\frac{1}{4}(0)\right)=\frac{3}{4}-\frac{1}{4}=\frac{1}{2}\]

Step 3: Find the x-coordinate of the centroid, \(\bar{x}\)

Next, find the x-coordinate (\(\bar{x}\)) by calculating the integral: \[ \bar{x} = \frac{1}{A}\int_0^1 x(\sqrt[3]{x} - x^3) dx = \frac{2}{1}\int_0^1 x(\sqrt[3]{x} - x^3) dx = 2\int_0^1 x(\sqrt[3]{x} - x^3) dx \] We calculate the integral: \[ 2\int_0^1 x(\sqrt[3]{x} - x^3) dx = 2\left[ \frac{3}{7}x^{7/3} - \frac{1}{8}x^8\right]_0^1 = 2\left(\frac{3}{7}(1)^{7/3} - \frac{1}{8}(1)^8 - (0)\right) =2\left(\frac{3}{7} - \frac{1}{8}\right) = \frac{61}{56} \]

Step 4: Find the y-coordinate of the centroid, \(\bar{y}\)

Finally, find the y-coordinate (\(\bar{y}\)) of the centroid: \[ \bar{y} = \frac{1}{A}\int_0^1 \frac{1}{2}(\sqrt[3]{x^2} - x^6) dx = 2\int_0^1\frac{1}{2}(\sqrt[3]{x^2} - x^6) dx = \int_0^1 (\sqrt[3]{x^2} - x^6) dx \] We calculate the integral: \[ \int_0^1 (\sqrt[3]{x^2} - x^6) dx = \left[ \frac{3}{5}x^{5/3} - \frac{1}{7}x^7 \right]_0^1 = \left(\frac{3}{5}(1)^{5/3} - \frac{1}{7}(1)^7 \right) - \left(\frac{3}{5}(0)^{5/3} - \frac{1}{7}(0)^7 \right) = \frac{16}{35} \]

Step 5: Find the centroid's coordinates

Now that we have the x-coordinate \(\bar{x}\) and the y-coordinate \(\bar{y}\) of the centroid, the centroid is given by the coordinates: \[(\bar{x}, \bar{y}) = \left(\frac{61}{56}, \frac{16}{35}\right) \] So the centroid of the region bounded by the graphs of the given equations is \(\left(\frac{61}{56}, \frac{16}{35}\right)\).

Key Concepts

Definite Integral

The concept of the definite integral is a cornerstone of calculus and serves as a tool for quantifying area under a curve. When we calculate a definite integral, we are essentially looking for the total accumulation of a quantity - in our context, the area - over a specific interval. Given a function f(x), the definite integral from a to b is represented by

\[ \int_a^b f(x) \, dx \]
where 'a' is the lower limit and 'b' is the upper limit of integration. To compute this integral, one typically relies on the Fundamental Theorem of Calculus, which connects antiderivatives with area. In the example provided, the definite integral is used to find both the area of the region and the moments necessary to compute the centroid, represented by specific integrals with the functions that bound the region in question.

Bounded Region

In calculus, a bounded region is an area confined within certain limits, usually by functions on a graph. These can be horizontal or vertical lines (like x=a or y=b) or more complex curves described by various functions. In our example, the region is enclosed by the functions \( y=x^{3} \), \( y=\sqrt[3]{x} \), \( x=0 \), and \( x=1 \). A key step in working with such a region is to first graph these functions and identify where they intersect, as this determines the limits of integration for calculating area and centroid.

Single Variable Calculus

Single variable calculus is the branch of mathematics that involves the study of rates of change (derivatives) and accumulation (integrals) of functions with one independent variable. It is essential for analyzing real-world phenomena where quantities vary continuously. The textbook exercise provided is an example of an application of single variable calculus, demonstrating how to find the centroid of a two-dimensional region. This involves both differentiation to obtain the equations of tangent lines or curves, and integration to consolidate the values between two points to find areas or centroids.

Area Calculation

Calculating the area under a curve or between curves is a fundamental application of integral calculus, referred to as area calculation. It is often visualized as finding the 'space' enclosed by graphed functions. In the example exercise, we use the definite integral to calculate the area (A) of the region bounded by the curves \( y=x^{3} \) and \( y=\sqrt[3]{x} \) between \( x=0 \) and \( x=1 \). This process involves integrating the difference between the two functions across the interval from 0 to 1. The result of this integral represents the total area of the region, which is essential for determining the centroid, as it acts as a scaling factor in the calculation.