Finding the Area Enclosed by a Curve and a Line

The problem asks us to find the area of the region enclosed by the curve \(y^2 = 2x\) and the straight line \(y = x\).

To find the area of the region enclosed by two curves, we typically follow these steps:

1. Find the points of intersection of the two curves.
2. Determine which curve is the upper boundary and which is the lower boundary within the region of interest.
3. Set up and evaluate the definite integral representing the area between the curves over the interval defined by the intersection points.

Step 1: Finding Intersection Points

We need to find the points where the curve \(y^2 = 2x\) and the line \(y = x\) intersect. We can substitute \(y = x\) into the equation of the curve:

\( (x)^2 = 2x \)

\( x^2 = 2x \)

Rearranging the equation to solve for \(x\):

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

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

This gives us two possible values for \(x\): \(x = 0\) or \(x = 2\).

Now, we find the corresponding \(y\) values using the equation of the line \(y = x\):

• If \(x = 0\), then \(y = 0\). The intersection point is (0, 0).
• If \(x = 2\), then \(y = 2\). The intersection point is (2, 2).

So, the two curves intersect at the points (0, 0) and (2, 2). These points define the limits of integration for finding the area of the enclosed region along the x-axis.

Step 2: Determining Upper and Lower Curves

The region is enclosed between \(x = 0\) and \(x = 2\). The curve \(y^2 = 2x\) can be written as \(y = \pm\sqrt{2x}\). Since the line \(y = x\) passes through the first quadrant (for \(x > 0\)), the enclosed region in the first quadrant is bounded by the line \(y = x\) and the upper half of the parabola \(y = \sqrt{2x}\).

To confirm which function is greater in the interval \(0 < x < 2\), let's pick a test point, say \(x = 1\):

• For the line \(y = x\), \(y = 1\).
• For the upper half of the parabola \(y = \sqrt{2x}\), \(y = \sqrt{2(1)} = \sqrt{2}\).

Since \(\sqrt{2} \approx 1.414\) and \(1.414 > 1\), the curve \(y = \sqrt{2x}\) is above the line \(y = x\) in the interval (0, 2).

Step 3: Setting Up the Integral

The area \(A\) between two curves \(y = f(x)\) and \(y = g(x)\) from \(x = a\) to \(x = b\), where \(f(x) \ge g(x)\) in the interval \([a, b]\), is given by the formula:

\( A = \int_{a}^{b} (f(x) - g(x)) \, dx \)

In our case, \(a = 0\), \(b = 2\), the upper curve is \(y = \sqrt{2x}\), and the lower curve is \(y = x\). So, the integral for the area is:

\( A = \int_{0}^{2} (\sqrt{2x} - x) \, dx \)

We can rewrite \(\sqrt{2x}\) as \(\sqrt{2} \cdot \sqrt{x} = \sqrt{2} x^{1/2}\).

\( A = \int_{0}^{2} (\sqrt{2} x^{1/2} - x) \, dx \)

Step 4: Evaluating the Integral

Now, we evaluate the definite integral:

\( A = \left[ \sqrt{2} \cdot \frac{x^{1/2+1}}{1/2+1} - \frac{x^{1+1}}{1+1} \right]_{0}^{2} \)

\( A = \left[ \sqrt{2} \cdot \frac{x^{3/2}}{3/2} - \frac{x^2}{2} \right]_{0}^{2} \)

\( A = \left[ \sqrt{2} \cdot \frac{2}{3} x^{3/2} - \frac{x^2}{2} \right]_{0}^{2} \)

\( A = \left[ \frac{2\sqrt{2}}{3} x^{3/2} - \frac{x^2}{2} \right]_{0}^{2} \)

Now, we apply the limits of integration:

\( A = \left( \frac{2\sqrt{2}}{3} (2)^{3/2} - \frac{(2)^2}{2} \right) - \left( \frac{2\sqrt{2}}{3} (0)^{3/2} - \frac{(0)^2}{2} \right) \)

Calculate the terms:

• \((2)^{3/2} = (2^1)(2^{1/2}) = 2\sqrt{2}\)
• \((2)^2 = 4\)

\( A = \left( \frac{2\sqrt{2}}{3} (2\sqrt{2}) - \frac{4}{2} \right) - (0 - 0) \)

\( A = \left( \frac{2 \cdot 2 \cdot \sqrt{2} \cdot \sqrt{2}}{3} - 2 \right) \)

\( A = \left( \frac{4 \cdot 2}{3} - 2 \right) \)

\( A = \left( \frac{8}{3} - 2 \right) \)

\( A = \frac{8}{3} - \frac{6}{3} \)

\( A = \frac{2}{3} \)

The area of the region enclosed by the curve \(y^2 = 2x\) and the line \(y = x\) is \(\frac{2}{3}\) square units.

Revision Table: Area Between Curves

This section summarizes the key steps and concepts for calculating the area between two functions.

Additional Information: Graphing and Area Calculation

Visualizing the region is very helpful when calculating the area between curves. The curve \(y^2 = 2x\) is a parabola opening to the right, symmetric about the x-axis. The line \(y = x\) is a straight line passing through the origin with a slope of 1.

When \(x \ge 0\), the parabola has an upper branch \(y = \sqrt{2x}\) and a lower branch \(y = -\sqrt{2x}\). The line \(y=x\) intersects the parabola where \(x=0\) (at (0,0)) and where \(x=2\) (at (2,2)). The region enclosed by the curve \(y^2=2x\) and the line \(y=x\) refers to the finite region bounded by both. This enclosed region is above the x-axis and is bounded by the upper branch of the parabola \(y=\sqrt{2x}\) and the line \(y=x\).

Alternatively, one could integrate with respect to \(y\). The equations are \(x = \frac{y^2}{2}\) and \(x = y\). Intersection points are (0,0) and (2,2), corresponding to \(y=0\) and \(y=2\). For \(y \in (0,2)\), the line \(x=y\) is to the right of the parabola \(x=\frac{y^2}{2}\) (e.g., at \(y=1\), \(x=1\) vs \(x=1/2\)). The area integral would be \(\int_{0}^{2} (y - \frac{y^2}{2}) \, dy\).

Let's quickly check this alternative integration:

\( \int_{0}^{2} (y - \frac{y^2}{2}) \, dy = \left[ \frac{y^2}{2} - \frac{y^3}{6} \right]_{0}^{2} \)

\( = \left( \frac{2^2}{2} - \frac{2^3}{6} \right) - (0 - 0) \)

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

\( = \left( 2 - \frac{4}{3} \right) \)

\( = \frac{6}{3} - \frac{4}{3} \)

\( = \frac{2}{3} \)

Both methods yield the same result, \(\frac{2}{3}\) square units, confirming the calculation.