Question

In Exercises \(41-44\), set up and evaluate the integrals for finding the area and moments about the \(x\) - and y-axes for the region bounded by the graphs of the equations. (Assume \(\rho=1\).) $$ y=2 x+4, y=0,0 \leq x \leq 3 $$

Short answer

The area under the curve is 21 square units. The first moment about the x-axis is 27 units and about the y-axis is 156 units.

Step-by-step solution

1. Find the Area Under the Line

The limits of integration are given by \(0 \leq x \leq 3\). The function to integrate is \(y=2x+4\). Therefore, the area under the line is given by \(A=\int_{0}^{3} y dx = \int_{0}^{3} (2x+4) dx\).

2. Evaluate the Integral for Area

Using rules for integration, the integral becomes \(A=[x^2+4x]_0^3\). Evaluating at the limits yields \(A=(3^2+4*3)- (0^2+4*0) = 21 square units\).

3. Find the Moments About the Axes

To find the first moments about the \(x\)- and \(y\)-axes, we need to evaluate the integrals \(\int x y dx\) and \(\int y^2 dx\) respectively. The moment about the x-axis ( \(M_x\) ) is given by \(\int_0^3 x (2x+4) dx\). Similarly, the moment about the y-axis ( \(M_y\) ) is given by \(\int_0^3 (2x+4)^2 dx\).

4. Evaluate the Integrals for Moments

Using the same rules for evaluating integrals, for \(M_x\) the integral becomes \(M_x=[\frac{x^3}{2}+2x^2]_0^3\) resulting in \(M_x=(9+18)-(0+0)=27 units\). For \(M_y\) the integral becomes \(M_y=[\frac{4x^3}{3}+8x^2+16x]_0^3\) resulting in \(M_y=(36+72+48)-(0+0+0)= 156 units\).

Key concepts

Integration

Integration is the mathematical process of finding the whole when you know only parts, typically represented by the area under a curve on a graph. It is a fundamental concept in calculus, utilized to solve a variety of real-world problems including calculating areas, volumes, and, as in the exercise, moments about axes. The basic idea is to break the area into infinitesimally small parts (think of them as incredibly thin rectangles), sum their areas, and take the limit as the width of these rectangles approaches zero. In the provided problem,

we need to integrate the function representing the line, which is given by the equation \( y = 2x + 4 \). To compute the area under this curve from \( x = 0 \) to \( x = 3 \), we set up an integral bounded by these limits. The area, \( A \), is then the result of the integral \( \int_{0}^{3} (2x+4) dx \). The integration process involves finding an antiderivative, a function whose derivative is the original function we are integrating, and then applying the Fundamental Theorem of Calculus which allows us to evaluate the integral using the antiderivative at the bounds of integration.

Area Under a Curve

The concept of area under a curve may seem quite literal at first—it's simply the area of the region that's beneath a curve and above the x-axis between two points on the horizontal axis. In the given exercise, this area corresponds to the region enclosed by the line \( y = 2x + 4 \), the x-axis (\( y=0 \)), and the given bounds on the x-axis from 0 to 3.

Mathematically, finding the area under a curve demonstrates integration in action. The integral of a function over a certain interval provides the cumulative sum of the areas of those infinitesimally thin rectangles mentioned earlier. When you evaluate this definite integral by plugging in the upper and lower bounds of the interval, you obtain the exact area of the region under the curve. This operation is pivotal in solving real-life problems where we need to calculate spaces enclosed by arbitrary shapes, such as land plots, lakes, or even the displacement of an object as a function of time in physics.

Moments About Axes

Moments about axes in calculus are measures of the distribution of the area's density across a given axis. Simply put, it's like trying to balance a cut-out shape on a pencil—if the shape is perfectly balanced, the point where it's balancing on the pencil is called the centroid. In a two-dimensional plane, moments are usually calculated about the x-axis and y-axis.

In the given problem, to calculate the first moment about the x-axis, represented as \( M_x \), we need to multiply each tiny slice of the area (each infinitesimally thin rectangle) by its distance from the x-axis, and then sum up all these products - this is done by the integral \( \int x y dx \). Conversely, for the moment about the y-axis, \( M_y \), we multiply each slice by its distance from the y-axis (it's horizontal this time, so we square the function) and then sum all these products using the integral \( \int y^2 dx \). These calculations help us understand where the center of mass of a given shape lies, which has all sorts of practical applications, from engineering to physics to even graphic design.