Calculating the Area Bounded by Parabolas

The problem asks for the area of the region enclosed by two parabolas given by the equations \(y^2 = 6(x - 1)\) and \(y^2 = 3x\). To find the area between these curves, we first need to determine their intersection points.

Finding the Intersection Points

We can find the intersection points by setting the expressions for \(y^2\) equal to each other:

\(6(x - 1) = 3x\)

\(6x - 6 = 3x\)

\(6x - 3x = 6\)

\(3x = 6\)

\(x = 2\)

Now substitute \(x = 2\) back into either equation to find the corresponding \(y\) values. Using \(y^2 = 3x\):

\(y^2 = 3(2)\)

\(y^2 = 6\)

\(y = \pm \sqrt{6}\)

The intersection points are \((2, \sqrt{6})\) and \((2, -\sqrt{6})\). This indicates that the region is bounded vertically by \(y = -\sqrt{6}\) and \(y = \sqrt{6}\).

Setting up the Area Integral

Since the equations are given in the form \(y^2 = f(x)\), the parabolas open to the right. It is easier to calculate the area by integrating with respect to \(y\). We need to express \(x\) in terms of \(y\) for both equations:

• From \(y^2 = 3x\), we get \(x_1 = \frac{y^2}{3}\).
• From \(y^2 = 6(x - 1)\), we get \(y^2 = 6x - 6\), which gives \(6x = y^2 + 6\), so \(x_2 = \frac{y^2 + 6}{6} = \frac{y^2}{6} + 1\).

To set up the integral \(\int (x_{\text{right}} - x_{\text{left}}) dy\), we need to determine which function represents the right boundary and which represents the left boundary between \(y = -\sqrt{6}\) and \(y = \sqrt{6}\). Let's pick a value of \(y\) within this range, say \(y=0\). For \(y=0\):

• \(x_1 = \frac{0^2}{3} = 0\) (for \(y^2 = 3x\)).
• \(x_2 = \frac{0^2}{6} + 1 = 1\) (for \(y^2 = 6(x - 1)\)).

Since \(1 > 0\), the parabola \(y^2 = 6(x - 1)\) (which is \(x = \frac{y^2}{6} + 1\)) is on the right of the parabola \(y^2 = 3x\) (which is \(x = \frac{y^2}{3}\)) in the region between the intersection points.

The area \(A\) is given by the definite integral from \(y = -\sqrt{6}\) to \(y = \sqrt{6}\) of the difference between the right curve and the left curve:

\(A = \int_{-\sqrt{6}}^{\sqrt{6}} \left( \left(\frac{y^2}{6} + 1\right) - \frac{y^2}{3} \right) dy\)

\(A = \int_{-\sqrt{6}}^{\sqrt{6}} \left( 1 + \frac{y^2}{6} - \frac{y^2}{3} \right) dy\)

\(A = \int_{-\sqrt{6}}^{\sqrt{6}} \left( 1 + \frac{y^2 - 2y^2}{6} \right) dy\)

\(A = \int_{-\sqrt{6}}^{\sqrt{6}} \left( 1 - \frac{y^2}{6} \right) dy\)

Evaluating the Integral

The integrand \(1 - \frac{y^2}{6}\) is an even function, and the limits of integration are symmetric about 0. We can use the property \(\int_{-a}^{a} f(y) dy = 2 \int_{0}^{a} f(y) dy\) for even functions.

\(A = 2 \int_{0}^{\sqrt{6}} \left( 1 - \frac{y^2}{6} \right) dy\)

Now, we find the antiderivative:

\(\int \left( 1 - \frac{y^2}{6} \right) dy = y - \frac{y^3}{18}\)

Evaluate the definite integral:

\(A = 2 \left[ y - \frac{y^3}{18} \right]_{0}^{\sqrt{6}}\)

\(A = 2 \left[ \left( \sqrt{6} - \frac{(\sqrt{6})^3}{18} \right) - \left( 0 - \frac{0^3}{18} \right) \right]\)

\(A = 2 \left[ \sqrt{6} - \frac{6\sqrt{6}}{18} \right]\)

\(A = 2 \left[ \sqrt{6} - \frac{\sqrt{6}}{3} \right]\)

Combine the terms inside the brackets:

\(A = 2 \left[ \frac{3\sqrt{6}}{3} - \frac{\sqrt{6}}{3} \right]\)

\(A = 2 \left[ \frac{3\sqrt{6} - \sqrt{6}}{3} \right]\)

\(A = 2 \left[ \frac{2\sqrt{6}}{3} \right]\)

\(A = \frac{4\sqrt{6}}{3}\)

The area of the region bounded by the two parabolas is \(\frac{4\sqrt{6}}{3}\) square units.

Revision Table: Area Calculation Steps

Additional Information: Area Between Curves

Finding the area between curves is a common application of definite integrals in calculus. The general approach depends on whether you integrate with respect to \(x\) or \(y\).

• Integration with respect to \(x\): If the region is bounded by curves \(y=f(x)\) and \(y=g(x)\) from \(x=a\) to \(x=b\), and \(f(x) \ge g(x)\) over the interval \([a, b]\), the area is given by \(\int_{a}^{b} (f(x) - g(x)) dx\).
• Integration with respect to \(y\): If the region is bounded by curves \(x=f(y)\) and \(x=g(y)\) from \(y=c\) to \(y=d\), and \(f(y) \ge g(y)\) over the interval \([c, d]\), the area is given by \(\int_{c}^{d} (f(y) - g(y)) dy\). Here, \(f(y)\) is the curve on the right and \(g(y)\) is the curve on the left.

In this problem, since the equations are given in the form \(y^2 = f(x)\) and the region is easily described by horizontal bounds (\(y=\pm\sqrt{6}\)), integrating with respect to \(y\) was the most straightforward method.