Question

Sketch the region bounded by the graphs of the given equations, show a typical slice, approximate its area, set up an integral, and calculate the area of the region. Make an estimate of the area to confirm your answer. \(x=-6 y^{2}+4 y, x+3 y-2=0\)

Short answer

The area of the region is approximately 0.463.

Step-by-step solution

Find Points of Intersection

To find the intersection points of the curves \( x = -6y^2 + 4y \) and \( x + 3y - 2 = 0 \), set the equations equal: \(-6y^2 + 4y = -3y + 2\). Simplifying gives the quadratic equation \(-6y^2 + 7y - 2 = 0\). Solving this quadratic using the quadratic formula yields the solutions \(y = \frac{1}{3}\) and \(y = 1\). Substitute these values back to find corresponding \(x\)-values: for \(y = \frac{1}{3}, x = -\frac{2}{3}\), and for \(y=1, x = -2\). Thus, points of intersection are \((-\frac{2}{3}, \frac{1}{3})\) and \((-2, 1)\).

Set Up the Integral for Area

We compute the area between the curves as the vertical slice extends from \(y = \frac{1}{3}\) to \(y = 1\). Set up the integral: the function to the right minus the function to the left, i.e., \[ \int_{y=\frac{1}{3}}^1 (3y - 2 - (-6y^2 + 4y)) \, dy. \] Simplifying gives \[ \int_{y=\frac{1}{3}}^1 (6y^2 - y - 2) \, dy. \]

Evaluate the Integral

Evaluate the integral \[ \int_{y=\frac{1}{3}}^1 (6y^2 - y - 2) \, dy \] by finding the antiderivative: \[ \left[ 2y^3 - \frac{y^2}{2} - 2y \right]_{\frac{1}{3}}^1. \] Calculate this by evaluating at the bounds: \( (2 \cdot 1^3 - \frac{1^2}{2} - 2 \cdot 1) - (2 \cdot (\frac{1}{3})^3 - \frac{(\frac{1}{3})^2}{2} - 2 \cdot \frac{1}{3}) \). This calculates to \(-0.5 - (-0.037037)\) which equals \(-0.462963\). Thus the area is \(0.462963\) (since area must be positive).

Estimate the Area for Confirmation

We estimate by calculating the simple geometric shapes within the bounds as the region is between two curves. Approximate using rectangles: approximately one rectangle of height \(\frac{2}{3}\) (the \(y\)-values difference) and average base\(\frac{2}{3}\), with simplified area equals approximately 0.5. This estimate confirms the calculated area is reasonable.

Key concepts

Area Between Curves

Finding the area between curves is an essential application of calculus integration. To comprehend this concept, imagine two curves plotted on a graph, and our goal is to find the region that lies between them. This region can often resemble a complex shape rather than simple geometric ones like rectangles or circles. Therefore, mathematicians use integration to accurately compute these areas.

• The goal is to determine the space enclosed by two curves.
• Use vertical or horizontal slices to evaluate these regions.
• Integrate the difference between the two functions over a specified interval.

To solve these problems specifically, set up an integral of the top function minus the bottom function over the shared interval. Calculating this integral gives you the area of the region bounded by the two curves.

Intersection Points

Finding intersection points is a fundamental step when calculating the area between curves. It indicates where the curves cross each other, serving as boundaries for integration.

• The intersection points help determine the limits for the definite integral.
• These points are usually where one function equals another.
• In the problem provided, we set the equations equal to each other to solve for values that both curves share.

First, solve the system of equations to find the recommended:

• Use algebraic methods like substitution or elimination, and sometimes graphical methods can be used for complex situations.

Rectify these points into the graph, ensuring proper understanding of how each function behaves.

Definite Integral Evaluation

Evaluating a definite integral is the process of determining the area under a curve within specific boundaries. This method is particularly effective when dealing with defined interval boundaries given by the intersection points.

• The definite integral takes the antiderivative over an interval.
• To calculate the definite integral, find the antiderivative and apply the Fundamental Theorem of Calculus.

Here's how you proceed:

• Determine the antiderivative of the function representing the area under the curve.
• Evaluate the antiderivative at the upper and lower bounds and subtract: \[ \left[ F(y) \right]_a^b = F(b) - F(a) \]
• This evaluation provides the exact area, although estimations can be useful for verification.

Quadratic Equations

Quadratic equations frequently appear in calculus, especially when dealing with curves described by polynomial equations of degree two. A quadratic equation has the general form \[-ax^2 + bx + c = 0\].

• It is characterized by a parabolic shape on a graph.
• Solving these equations involves techniques such as factoring, completing the square, or using the quadratic formula.

In the context of our example:

• The intersection and integration involve solving a quadratic equation to locate specific y-values.
• Employ the quadratic formula, \[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\], to find where curves intersect.

Understanding solving quadratic equations allows for pinpointing critical points, which aids in setting integration boundaries effectively.