Problem 23 - Single Variable Calculus

Question

Find the centroid of the region bounded by the graphs of the given equations. $$ y=6-x^{2}, \quad y=3-2 x $$

Step-by-Step Solution

Answer

The centroid of the region bounded by the graphs of the given equations \(y=6-x^{2}\) and \(y=3-2x\) is approximately at \((0.708, 1.667)\).

Step 1: Sketch the Graphs of the Functions

First, we need to draw the graphs of both functions: $$y=6-x^2$$ and $$y=3-2x$$. The first equation represents a downward-opening parabola with vertex (0,6). The second equation is a linear function with slope -2 and y-intercept 3.

Step 2: Determine the Intersection Points

Now, let's find the intersection points of the two functions by solving the system of equations: $$ y = 6-x^2 \\ y = 3-2x $$ Set the equations equal: $$6-x^2 = 3-2x$$ Solve for x: $$x^2 - 2x - 3 = 0$$ Factoring: $$(x-3)(x+1) = 0$$ So, x = 3 and x = -1. To find the corresponding y-values, plug these values back into either of the original equations (we'll use the second one): For x = 3: $$y = 3-2(3) = -3$$ For x = -1: $$y = 3-2(-1) = 5$$ The intersection points are (3, -3) and (-1, 5).

Step 3: Find the Area of the Bounded Region

Now, we will find the area of the region bounded by the given functions by integrating the difference of the functions (higher minus the lower function) over the interval from the left to the right intersection point: $$Area = \int_{-1}^{3} [(3-2x) - (6-x^2)] dx$$ This simplifies to: $$Area = \int_{-1}^{3} (-x^2 + 2x -3) dx$$ Integrate the function and evaluate the result: $$Area = \left[ -\frac{1}{3}x^3 + x^2 -3x \right]_{-1}^{3}$$ $$Area = \left(-\frac{1}{3}(3)^3 + (3)^2 -3(3)\right) - \left(-\frac{1}{3}(-1)^3 + (-1)^2 -3(-1)\right) $$ $$Area = (-9 + 9 - 9) - (\frac{1}{3} + 1 + 3)$$ $$Area = 8$$ The area of the region is 8 square units.

Step 4: Calculate the Coordinates of the Centroid

Finally, we will find the coordinates of the centroid (x̅,y̅) using the formulas: $$x̅ = \frac{1}{Area}\int_{-1}^{3} x[(3-2x) - (6-x^2)] dx$$ $$y̅ = \frac{1}{2Area}\int_{-1}^{3} [(3-2x)^2 - (6-x^2)^2] dx$$ Calculate the x-coordinate of the centroid: $$x̅ = \frac{1}{8}\int_{-1}^{3} x(-x^2 + 2x -3) dx$$ $$x̅ = \frac{1}{8}\left[ -\frac{1}{4}x^4 + \frac{1}{3}x^3 -\frac{3}{2}x^2\right]_{-1}^{3}$$ $$x̅ = \frac{1}{8}( -\frac{81}{4} + 9 - \frac{27}{2}) - (-\frac{1}{4} + \frac{1}{3} -\frac{1}{2})$$ $$x̅ = \frac{1}{8}(- \frac{21}{4} + \frac{2}{3} + \frac{1}{2})$$ $$x̅ = \frac{17}{24}$$ Now, calculate the y-coordinate of the centroid: $$y̅ = \frac{1}{16}\int_{-1}^{3} [(3-2x)^2 - (6-x^2)^2] dx$$ $$y̅ = \frac{1}{16}\left[-\frac{1}{3}x^6 + 22x^4 - 72x^2 \right]_{-1}^{3}$$ $$y̅ = \frac{1}{16}( -\frac{1}{3}(3)^6 + 22(3)^4 - 72(3)^2 - (-\frac{1}{3}(-1)^6 + 22(-1)^4 - 72(-1)^2 ))$$ $$y̅ = \frac{5}{3}$$ The centroid of the region is approximately at (0.708, 1.667).

Key Concepts

Intersection Points

Finding the intersection points of two equations is a crucial step in understanding how these lines or curves interact with each other. In the context of coordinate geometry, an intersection point is where two graphs meet. For our equations, \(y = 6-x^2\) and \(y = 3-2x\), we find the intersection by setting them equal to each other, as both share the same \(y\)-value at these points. This leads to solving:

• \(6-x^2 = 3-2x\) simplifies to \(x^2 - 2x - 3 = 0\)
• Factoring gives \((x-3)(x+1) = 0\)
• This results in \(x\) values of \(3\) and \(-1\)

By substituting these \(x\) values back into one of the original equations \(y = 3-2x\), we find the corresponding \(y\)-coordinates: for \(x = 3\), \(y = -3\) and for \(x = -1\), \(y = 5\). So, the intersection points are \((3, -3)\) and \((-1, 5)\). These points help us define the bounded region we are analyzing.

Integral Calculus

Integral calculus helps us handle quantities where changes occur over an interval, which is perfect for finding areas and centroids. When calculating areas or centroids involving curves, we use definite integrals. Here, the integral finds the area between the curves \((3 - 2x)\) and \((6 - x^2)\). This is achieved by computing:

• \(\int_{-1}^{3} [(3-2x) - (6-x^2)] \, dx\)
• The integral simplifies to \(\int_{-1}^{3} (-x^2 + 2x -3) \, dx\)

We evaluate the definite integral between the x-values of our intersection points (-1 and 3). The integral is worked out by calculating the antiderivative:

• \(-\frac{1}{3}x^3 + x^2 -3x\)
• Then, substituting the limits to find the effective area.

Integral calculus not only aids in finding areas but is also vital for finding centroids, as it allows us to evenly distribute mass across a shape.

Area of Bounded Region

The area of a bounded region refers to the enclosed space within intersecting curves or lines. Here, this area is essential when calculating the centroid since the centroid's position depends on the region's distribution. To find the area between two curves, we use the integral of the difference of the functions:

• Set up the integral: \(\int_{-1}^{3} [(3-2x) - (6-x^2)] \, dx\)
• Simplify the functions to \(-x^2 + 2x -3\)
• Find the antiderivative and evaluate the limits: \(-\frac{1}{3}x^3 + x^2 -3x\)
• Substitute limits \(-1\) and \(3\) to calculate area \((Area = 8)\)

Calculating the area helps paint a complete picture of the region that we analyze for further geometric properties, like finding centroids or centers of mass.

Coordinate Geometry

Coordinate geometry permits us to place geometric figures within a plane, using a coordinate system to identify each point. This is essential for locating intersection points and defining the region of interest. This particular problem involves a straight line and a parabola. By graphing these, one can visualize their interaction:

• \(y = 6-x^2\): A parabola opening downwards with a vertex at \((0,6)\)
• \(y = 3-2x\): A line with a slope of \(-2\) crossing the y-axis at \(3\)

Using algebra and calculus together, we utilize coordinate geometry to:

• Find exact intersection points, allowing us to use integral calculus effectively.
• Provide the graphical basis needed for integral setup and calculation.

In essence, coordinate geometry translates graphical data into solvable equations, unveiling valuable details about geometric relationships.