Problem 8 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

The centroid is at \( \left( \frac{2}{3}, \frac{2}{3} \right) \).

Step 1: Understand the Region

We have three boundaries: the line \( y = 2 - x \), the line \( y = 0 \) (the x-axis), and the line \( x = 0 \) (the y-axis). The region is a right triangle with vertices at \( (0, 0) \), \( (2, 0) \), and \( (0, 2) \).

Step 2: Determine the Area of the Triangle

The base of the triangle along the x-axis is \(2\) units, and the height along the y-axis is also \(2\) units. The area \( A \) of the triangle is given by the formula \( A = \frac{1}{2} \times ext{base} \times ext{height} = \frac{1}{2} \times 2 \times 2 = 2 \).

Step 3: Use Symmetry to Find the Centroid

The region is a right triangle and is symmetric about the line \( x = y \). The centroid of a right triangle can be found at \( \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \), where \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \) are the vertices of the triangle.

Step 4: Calculate the Centroid Coordinates

Substitute the coordinates of the vertices \( (0, 0), (2, 0), (0, 2) \) into the centroid formula:\[\left( \frac{0 + 2 + 0}{3}, \frac{0 + 0 + 2}{3} \right) = \left( \frac{2}{3}, \frac{2}{3} \right)\]

Step 5: Verify and Conclusion

The calculated centroid \( \left( \frac{2}{3}, \frac{2}{3} \right) \) is consistent with the symmetry of the triangle. The centroid should lie on the line \( x = y \), and this point satisfies that condition, confirming it is correct.

Key Concepts

Coordinate Geometry

Coordinate geometry, or analytic geometry, bridges the gap between algebra and geometry by using coordinates to study the properties and positions of geometric figures. This approach makes it easy to understand various geometric concepts using equations. In coordinate geometry, each point in space is described by ordered pairs \(x, y\) on a two-dimensional plane known as the Cartesian plane. In the exercise, the region is determined by the intersection of lines given by equations. Here, we have three lines: \(y = 2 - x\), \(y = 0\) (x-axis), and \(x = 0\) (y-axis). Using coordinate geometry, we identify the triangular region with vertices at (0,0), (2,0), and (0,2). By sketching these lines, we can visualize the shape and determine the included area. This visual understanding is key to finding properties like the centroid efficiently.

Triangle Symmetry

Symmetry simplifies many mathematical problems, especially in geometry. When we say a shape is symmetric, it can be divided into parts that are mirror images of each other. For triangles, symmetry can help in predicting features like location of its centroid.In our exercise, the triangle formed by the region bounded by \(y = 2-x\), \(y=0\), and \(x=0\) is a right triangle, which is symmetric about the line \(x = y\). This means if you fold the triangle along the line \(x = y\), both halves will match perfectly. The symmetry in this scenario allows us to predict that the centroid - the point where the triangle's medians intersect - should lie on this line. This insight simplifies finding the centroid's location.

Centroid Formula

The centroid of a triangle is the point where its three medians intersect. A median is a line segment joining a vertex to the midpoint of the opposite side. The centroid is also known as the triangle's "center of gravity" because it is the balance point of the triangle.The centroid formula for determining the centroid's coordinates \((x_c, y_c)\) is:

• For the x-coordinate: \((x_1 + x_2 + x_3)/3\)
• For the y-coordinate: \((y_1 + y_2 + y_3)/3\)

By substituting the triangular vertices \( (0,0), (2,0), (0,2) \) into these formulas, we find the centroid coordinates:\[ \left(\frac{0+2+0}{3}, \frac{0+0+2}{3}\right) = \left( \frac{2}{3}, \frac{2}{3} \right)\]This calculated centroid \(\left(\frac{2}{3},\frac{2}{3}\right)\) lies on the symmetry line \(x = y\), verifying its correctness. This step shows how geometric properties and algebra work hand in hand to provide simple and reliable solutions.