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Chapter 14: Problem 4

Find the centroid of the triangular region cut from the first quadrant by the line \(x+y=3\)

Short Answer

Expert verified
The centroid of the triangular region is at the point (1, 1).

Step by step solution

01

Determine the Vertices of the Triangle

The triangle is bounded by the line \(x+y=3\) in the first quadrant. To find the vertices, consider the intersections with the axes. Set \(x=0\), then \(y=3\), giving us the point \((0,3)\). Similarly, set \(y=0\), then \(x=3\), giving the point \((3,0)\). The third vertex is the origin \((0,0)\), as the line cuts the triangle off in the first quadrant.
02

Use the Centroid Formula

The formula for the centroid \((x_c, y_c)\) of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) is \(x_c = \frac{x_1 + x_2 + x_3}{3}\) and \(y_c = \frac{y_1 + y_2 + y_3}{3}\). Substitute the vertices \((0,0)\), \((3,0)\), and \((0,3)\) into these equations.
03

Calculate the Centroid Coordinates

Substitute the vertex coordinates into the centroid formula: \[x_c = \frac{0 + 3 + 0}{3} = 1\] and \[y_c = \frac{0 + 0 + 3}{3} = 1\]. So, the centroid of the triangle is at \((1, 1)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

First Quadrant
The coordinate plane is divided into four quadrants. Each quadrant is determined by the positive and negative values of the x and y axes. The first quadrant is particularly special because both the x and y coordinates are positive.
It's represented by the upper right section of the coordinate system.
This is important when dealing with geometrical figures, as it defines which section of the plane the figure lies in. In this exercise, the triangle is formed by the line equation and the axes, all within the first quadrant.
This means the vertices and centroid of the triangle have coordinates that are all positive.
Centroid Formula
The centroid of a triangle is the point where its three medians intersect, and it serves as a balancing point or center of mass of the triangle.
Mathematically, it can be found using the centroid formula.
Given a triangle with vertices
  • a o coordinate of \((x_1, y_1)\)
  • a b coordinate of \((x_2, y_2)\)
  • a c coordinate of \((x_3, y_3)\)
The coordinates of the centroid \((x_c, y_c)\) are given by \(x_c = \frac{x_1 + x_2 + x_3}{3}\) and \(y_c = \frac{y_1 + y_2 + y_3}{3}\).
This formula averages the x and y coordinates of the triangle's vertices, giving a point inside the triangle that is like its center.
Triangular Region
A triangular region on the coordinate plane refers to the area enclosed by the triangle. In this case, the triangular region is defined by the vertices determined by the line, the x-axis, and the y-axis, all of which lie in the first quadrant.
The vertices of this specific triangular region are at \((0,0)\), \((3,0)\), and \((0,3)\).
鈥� The point \((0,0)\) denotes the origin, where both x and y are zero.鈥� The point \((3,0)\) is where the line intersects the x-axis.鈥� The point \((0,3)\) is where the line crosses the y-axis.
Understanding the vertices and how the triangle is formed helps in visualizing how to apply the centroid formula and what results to expect.
Coordinate Geometry
Coordinate geometry, also known as analytic geometry, combines algebra and geometry using the coordinate plane. It's a helpful tool in solving geometric problems by representing figures with equations.
In this exercise, a line equation \(x + y = 3\) is used to determine the vertices of a triangle within the first quadrant and to calculate its centroid.
Here are a few applications:
  • Locating points: Helps find intersection points on axes, like finding vertices of the triangle.
  • Using formulas: Utilizes formulas like the centroid formula to find central points based on given vertices coordinates.
  • Graphing shapes: Allows you to draw and understand the spatial relationship between figures on the plane.
Coordinate geometry is a fundamental concept, connecting algebraic equations to geometric figures, essential for tackling problems involving shapes and figures on a plane.

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Most popular questions from this chapter

Find the volume of the solid whose base is the region in the \(x y\) -plane that is bounded by the parabola \(y=4-x^{2}\) and the line \(y=3 x,\) while the top of the solid is bounded by the plane \(z=x+4\)

Let \(D\) be the smaller cap cut from a solid ball of radius 2 units by a plane 1 unit from the center of the sphere. Express the volume of \(D\) as an iterated triple integral in (a) spherical, (b) cylindrical, and (c) rectangular coordinates. Then (d) find the volume by evaluating one of the three triple integrals.

a. Center of mass Find the center of mass of a solid of constant density bounded below by the paraboloid \(z=x^{2}+y^{2}\) and above by the plane \(z=4\) b. Find the plane \(z=c\) that divides the solid into two parts of equal volume. This plane does not pass through the center of mass.

The integrals we have seen so far suggest that there are preferred orders of integration for cylindrical coordinates, but other orders usually work well and are occasionally easier to evaluate. Evaluate the integrals. $$\int_{0}^{2} \int_{r-2}^{\sqrt{4-r^{2}}} \int_{0}^{2 \pi}(r \sin \theta+1) r d \theta d z d r$$

Centroid of a solid semiellipsoid Assuming the result that the centroid of a solid hemisphere lies on the axis of symmetry threeeighths of the way from the base toward the top, show, by transforming the appropriate integrals, that the center of mass of a solid semiellipsoid \(\left(x^{2} / a^{2}\right)+\left(y^{2} / b^{2}\right)+\left(z^{2} / c^{2}\right) \leq 1, z \geq 0,\) lies on the z-axis three-eighths of the way from the base toward the top. (You can do this without evaluating any of the integrals.)
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