Problem 14 - Single Variable Calculus

Question

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

Step-by-Step Solution

Answer

The centroid of the region bounded by the graphs of \(y=x^3\), \(y=0\), and \(x=3\) is \((\frac{972}{405}, \frac{4374}{567})\).

Step 1: Find the area of the region

To find the area of the region, integrate \(y = x^3\) with respect to \(x\) from \(x = 0\) to \(x = 3\): $$ A = \int_{0}^{3} x^3 dx $$

Step 2: Evaluate the integral to find the area

Use the power rule to evaluate the integral: $$ A = \frac{1}{4} x^4 \Big|_{0}^{3} = \frac{1}{4} (3^4 - 0^4) = \frac{81}{4} $$ So, the area of the region is \(\frac{81}{4}\).

Step 3: Find the first moment about the x-axis

To find the first moment about the x-axis, integrate the product of the function with x over the same interval, and divide the result by the total area: $$ M_x = \frac{1}{A} \int_{0}^{3} x \cdot x^3 dx = \frac{1}{\frac{81}{4}} \int_{0}^{3} x^4 dx $$

Step 4: Evaluate the integral for the first moment about the x-axis

Use the power rule to evaluate the integral: $$ M_x = \frac{1}{\frac{81}{4}} \cdot \frac{1}{5} x^5 \Big|_{0}^{3} = \frac{1}{\frac{81}{4}} \cdot \frac{1}{5} (3^5 - 0^5) = \frac{4}{405} (243) $$ So, the first moment about the x-axis is \(\frac{972}{405}\).

Step 5: Find the first moment about the y-axis

To find the first moment about the y-axis, integrate the product of the function with y over the same interval, and divide the result by the total area: $$ M_y = \frac{1}{A} \int_{0}^{3} (x^3) \cdot x^3 dx = \frac{1}{\frac{81}{4}} \int_{0}^{3} x^6 dx $$

Step 6: Evaluate the integral for the first moment about the y-axis

Use the power rule to evaluate the integral: $$ M_y = \frac{1}{\frac{81}{4}} \cdot \frac{1}{7} x^7 \Big|_{0}^{3} = \frac{1}{\frac{81}{4}} \cdot \frac{1}{7} (3^7 - 0^7) = \frac{4}{567} (2187) $$ So, the first moment about the y-axis is \(\frac{4374}{567}\).

Step 7: Calculate the centroid coordinates

The centroid's coordinates are given by \((M_x, M_y)\): $$ (\frac{972}{405}, \frac{4374}{567}) $$ So, the centroid of the region is \((\frac{972}{405}, \frac{4374}{567})\).

Key Concepts

Definite Integral Calculus

Definite integral calculus is a fundamental concept in mathematics that calculates the accumulation of quantities, such as areas under curves. Imagine painting a floor; the amount of paint used can be considered analogous to the area covered, and calculus helps us figure out just how much paint we need for oddly shaped spaces. In the exercise provided, finding the centroid of a region involving the curve y = x^3, the definite integral comes in handy to calculate the area (A) under the curve from x = 0 to x = 3. To do so, the integral of x^3 with respect to x is evaluated over this interval.

When solving real-world problems or homework exercises, it's not just about finding the area. The process also includes understanding the impact of the integrand's shape and the limits of integration. It requires careful execution of other calculus rules, such as the power rule for integration, to arrive at the correct solution.

Moment of a Region

Moments in physics and mathematics represent the weighted contributions of each point in a region or distribution. The moment of a region can be likened to finding the balance point of a seesaw. For a region in the plane, the first moments about the x- and y-axes (M_x and M_y) are crucial for locating the centroid, which is the geometric center like the 'sweet spot' where a flat cutout of the region could balance perfectly.

In our exercise, the moments are found by integrating the distance of each infinitesimal area piece to an axis, multiplied by the density (here, represented by the function y = x^3), over the region of interest. The results give us values that reflect how the shape's mass is distributed around each axis, guiding us to the centroid.

Power Rule Integration

Power rule integration simplifies the process of integration when dealing with polynomials. The rule states: to integrate a power x^n, increase the exponent by one and divide by the new exponent. In algebra, think of it similar to climbing up a ladder; as you take a step up, you also stretch out to maintain proportion.

In the context of our exercise, when we integrate x^3, we apply the power rule to get x^4/4. It's like a shortcut in calculus. Just like following a well-trodden path through a forest, power rule integration gives us a clear and efficient route to the solution, without getting tangled in complications that could arise from a more brute force method.

Single Variable Calculus

Single variable calculus deals with functions of one variable and encompasses all the operations that can be performed on these functions, including differentiation and integration. Think of it as examining a one-dimensional path through the vast mathematical landscape. You can move forward or backward, speed up or slow down, but you cannot veer off to the side.

In our exercise, we focus on a specific path—y = x^3—over an interval from 0 to 3 to find the region's area and moments. All of these calculations are based on the principles of single variable calculus, and they illustrate how even a single, one-dimensional input can yield rich and complex results that inform us about two-dimensional geometric properties.