Question

Find the centroid of the region bounded by the given curves. Make a sketch and use symmetry where possible. $$ y=2-x, y=0, x=0 $$

Short answer

The centroid of the region is \( \left( \frac{2}{3}, \frac{2}{3} \right) \).

Step-by-step solution

Understand the Region Bounded by the Curves

The given curves are \( y = 2 - x \), \( y = 0 \), and \( x = 0 \). The region bounded by these curves is a right triangle. This triangle is formed by the line \( y = 2 - x \) (a straight line with y-intercept 2 and slope -1), the horizontal line \( y = 0 \) (x-axis), and the vertical line \( x = 0 \) (y-axis).

Sketch the Region

Draw the x-y coordinate plane. The line \( y = 2 - x \) intersects the y-axis at (0, 2) and the x-axis at (2, 0). The region of interest is the triangular region above the x-axis and to the right of the y-axis, bounded by these two points.

Use Symmetry to Simplify Calculations

Notice that the triangle has a vertical line of symmetry \( x = 1 \) through its midpoint between (0,2) and (2,0). This symmetry indicates that the centroid will lie along this line, which means the x-coordinate of the centroid is \( \frac{1}{2}(2 + 0) = 1 \).

Calculate the Area of the Triangle

The area \( A \) of the triangle is given by: \[ A = \frac{1}{2} \times base \times height = \frac{1}{2} \times 2 \times 2 = 2. \]

Mathematical Formula for the Centroid

The centroid \((\overline{x}, \overline{y})\) of a region bounded by curves is given by: \[ \overline{x} = \frac{1}{A} \int_{a}^{b} x [f(x) - g(x)] \, dx, \]\[ \overline{y} = \frac{1}{2A} \int_{a}^{b} [f(x)^2 - g(x)^2] \, dx. \] Here, \( f(x) = 2 - x \) and \( g(x) = 0 \).

Calculate \( \overline{x} \) Coordinate

Evaluate the x-coordinate of the centroid:\[ \overline{x} = \frac{1}{2} \int_0^2 x (2 - x) \, dx = \frac{1}{2} \left[ \int_0^2 (2x - x^2) \, dx \right]. \]First, integrate \( 2x \and -x^2 \) separately:\[ \int 2x \, dx = x^2 \Big|_0^2 = 4, \]\[ \int x^2 \, dx = \frac{x^3}{3} \Big|_0^2 = \frac{8}{3}. \]Thus, \[ \overline{x} = \frac{1}{2} \left( 4 - \frac{8}{3} \right) = \frac{1}{2} \times \frac{4}{3} = \frac{2}{3}. \]

Calculate \( \overline{y} \) Coordinate

Evaluate the y-coordinate of the centroid:\[ \overline{y} = \frac{1}{4} \int_0^2 (4 - 4x + x^2) \, dx. \] Integrate each term separately:\[ \int 4 \, dx = 4x \Big|_0^2 = 8, \]\[ \int 4x \, dx = 2x^2 \Big|_0^2 = 8, \]\[ \int x^2 \, dx = \frac{x^3}{3} \Big|_0^2 = \frac{8}{3}. \]Then, substitute back:\[ \overline{y} = \frac{1}{4} (8 - 8 + \frac{8}{3}) = \frac{2}{3}. \]

Combine the Coordinates

The centroid of the triangular region is \( (\overline{x}, \overline{y}) = \left( \frac{2}{3}, \frac{2}{3} \right)\).

Key concepts

Bounded regions in calculus

In calculus, a bounded region is an area enclosed by curves, which defines a geometric shape on the coordinate plane. Understanding these regions is crucial, especially when calculating properties like the centroid. In our exercise, the region is bounded by the curves \(y = 2-x\), \(y = 0\), and \(x = 0\). This forms a right triangle in the first quadrant of the coordinate plane.

The boundaries include the horizontal line \(y=0\), which is simply the x-axis, and \(x=0\), the y-axis. The line \(y=2-x\) is a straight line with a slope of -1 that decreases from the point (0,2) on the y-axis to (2,0) on the x-axis. The area of interest lies between these intersecting lines, resulting in the triangular region.

Sketching the bounded region can make it easier to apply your calculus knowledge, such as setting up integrals or identifying symmetry to simplify calculations. Ensuring accuracy in understanding and sketching these regions will aid in correctly determining the properties of the shape, such as the centroid.

Integral calculus

Integral calculus provides tools for finding the areas of regions with unknown or irregular shapes. Calculating the centroid, or the geometric center, of an area involves solving definite integrals to locate the mean x and y values of the shape. Here's how we do it in our exercise.

To find the x-coordinate of the centroid \((\overline{x})\), you use the formula:

• \( \overline{x} = \frac{1}{A} \int_{a}^{b} x [f(x) - g(x)] \, dx \)

Here, \(A\) is the area of the region, and \(f(x)\) and \(g(x)\) are the functions describing the boundary curves. In our example, \(f(x) = 2-x\) and \(g(x) = 0\). We integrate from 0 to 2, the limits of our bounded region.

Next, the y-coordinate \((\overline{y})\) uses:

• \( \overline{y} = \frac{1}{2A} \int_{a}^{b} [f(x)^2 - g(x)^2] \, dx \)

This process involves integrating the square of the boundary function, quickly revealing the vertical mean position.

These formulas help determine the centroid's position, ensuring it reflects the true geometric center of the triangular region.

Symmetry in geometry

Symmetry in geometry involves identifying a specific line or point where a shape is a mirror image of itself. Recognizing symmetry can drastically simplify problems in calculus by reducing the complexity of certain calculations. In this exercise, recognizing symmetry played a crucial role in finding the centroid.

For the triangular region bounded by \(y = 2-x\), \(y=0\), and \(x=0\), there's vertical symmetry about the line \(x = 1\). This is the midpoint between the line's intercepts at (0, 2) and (2, 0). Due to this symmetry, the x-coordinate of the centroid aligns along this line, simplifying our task to just calculate the y-coordinate using integration.

Utilizing symmetry makes finding centroids more efficient in geometry and physics since half of the work is often intuitive through the symmetry property. This mastery is especially useful in more complicated shapes where symmetric properties are less obvious but equally present. When you see symmetry in a problem, it's not just a detail; it’s an effective tool to refine your mathematical approach.