Calculating the Area Enclosed by a Parabola and a Line

We need to find the area bounded by the parabola given by the equation \( x^2 = 6y \) and the line given by the equation \( x - 6y + 2 = 0 \).

Finding the Points of Intersection

To find the area enclosed by the two curves, we first need to determine the points where they intersect. We can do this by solving the two equations simultaneously.

The equations are:

1. Parabola: \( x^2 = 6y \)
2. Line: \( x - 6y + 2 = 0 \)

From the first equation, we can see that \( 6y \) is equal to \( x^2 \). We can substitute this expression for \( 6y \) into the second equation:

\( x - (x^2) + 2 = 0 \)

Rearranging the terms to form a standard quadratic equation:

\( -x^2 + x + 2 = 0 \)

Multiplying by -1 to make the \( x^2 \) coefficient positive:

\( x^2 - x - 2 = 0 \)

Now, we can factor this quadratic equation to find the values of \( x \) at the intersection points:

\( (x - 2)(x + 1) = 0 \)

This gives us two possible values for \( x \):

• \( x - 2 = 0 \Rightarrow x = 2 \)
• \( x + 1 = 0 \Rightarrow x = -1 \)

These x-values, \( x = -1 \) and \( x = 2 \), are the limits of integration for finding the area enclosed by the parabola and the line.

Expressing the Equations as Functions of x

To integrate with respect to \( x \), we need to express both the parabola and the line equations in the form \( y = f(x) \).

For the parabola \( x^2 = 6y \):

\( y_p = \frac{x^2}{6} \)

For the line \( x - 6y + 2 = 0 \):

\( 6y = x + 2 \)

\( y_l = \frac{x + 2}{6} \)

Determining the Upper and Lower Curves

Between the intersection points \( x = -1 \) and \( x = 2 \), we need to determine which curve is above the other. We can pick a test point within this interval, for example, \( x = 0 \).

For the line: \( y_l(0) = \frac{0 + 2}{6} = \frac{2}{6} = \frac{1}{3} \)

For the parabola: \( y_p(0) = \frac{0^2}{6} = 0 \)

Since \( \frac{1}{3} > 0 \), the line \( y_l = \frac{x + 2}{6} \) is above the parabola \( y_p = \frac{x^2}{6} \) in the interval \( [-1, 2] \).

Setting up the Definite Integral for Area

The area enclosed by the two curves is given by the definite integral of the difference between the upper curve and the lower curve, from the lower x-limit to the upper x-limit:

\( \text{Area} = \int_{a}^{b} (y_{\text{upper}} - y_{\text{lower}}) \, dx \)

In our case, \( a = -1 \), \( b = 2 \), \( y_{\text{upper}} = \frac{x + 2}{6} \), and \( y_{\text{lower}} = \frac{x^2}{6} \).

\( \text{Area} = \int_{-1}^{2} \left( \frac{x + 2}{6} - \frac{x^2}{6} \right) \, dx \)

We can factor out the constant \( \frac{1}{6} \):

\( \text{Area} = \frac{1}{6} \int_{-1}^{2} (x + 2 - x^2) \, dx \)

Evaluating the Definite Integral

Now, we evaluate the integral:

\( \int (x + 2 - x^2) \, dx = \frac{x^2}{2} + 2x - \frac{x^3}{3} \)

Evaluate this expression at the upper limit (x=2) and the lower limit (x=-1):

At \( x = 2 \):

\( \left( \frac{2^2}{2} + 2(2) - \frac{2^3}{3} \right) = \left( \frac{4}{2} + 4 - \frac{8}{3} \right) = \left( 2 + 4 - \frac{8}{3} \right) = \left( 6 - \frac{8}{3} \right) \)

To combine, find a common denominator (3): \( 6 - \frac{8}{3} = \frac{18}{3} - \frac{8}{3} = \frac{10}{3} \)

At \( x = -1 \):

\( \left( \frac{(-1)^2}{2} + 2(-1) - \frac{(-1)^3}{3} \right) = \left( \frac{1}{2} - 2 - \frac{-1}{3} \right) = \left( \frac{1}{2} - 2 + \frac{1}{3} \right) \)

To combine, find a common denominator (6): \( \frac{3}{6} - \frac{12}{6} + \frac{2}{6} = \frac{3 - 12 + 2}{6} = \frac{-7}{6} \)

Now subtract the value at the lower limit from the value at the upper limit:

\( \left( \frac{10}{3} \right) - \left( \frac{-7}{6} \right) = \frac{10}{3} + \frac{7}{6} \)

Find a common denominator (6): \( \frac{20}{6} + \frac{7}{6} = \frac{27}{6} \)

Calculating the Final Area

Finally, multiply the result of the definite integral by the factor \( \frac{1}{6} \):

\( \text{Area} = \frac{1}{6} \times \frac{27}{6} = \frac{27}{36} \)

Simplify the fraction:

\( \frac{27}{36} = \frac{9 \times 3}{9 \times 4} = \frac{3}{4} \)

The area enclosed by the parabola \( x^2 = 6y \) and the line \( x - 6y + 2 = 0 \) is \( \frac{3}{4} \) square units.

Revision Table: Area Calculation Concepts

Additional Information: Parabola and Line Geometry

A parabola is a U-shaped curve defined by a quadratic equation. The equation \( x^2 = 6y \) represents a parabola that opens upwards, symmetric about the y-axis, with its vertex at the origin (0,0).

A linear equation like \( x - 6y + 2 = 0 \) represents a straight line. The intersection points are where the parabola and the line meet. These points define a finite region whose area can be calculated using integration.

Understanding the graphs of these functions can help visualize the enclosed region and confirm which function is above the other in the interval of interest. For \( x^2 = 6y \), \( y = x^2/6 \), and for \( x - 6y + 2 = 0 \), \( y = (x+2)/6 \). Between \( x=-1 \) and \( x=2 \), the line \( y = (x+2)/6 \) is indeed above the parabola \( y = x^2/6 \).