Finding the Area Enclosed Between Curves

This problem asks for the calculation of the area of the region enclosed between two functions: the parabola defined by \(y^2 = 2x\) and the straight line defined by \(y = x\). To solve this, we need to find the intersection points of these curves and then set up and evaluate a definite integral.

Step 1: Identify Intersection Points

We first determine where the curve \(y^2 = 2x\) and the line \(y = x\) intersect. We can substitute \(y=x\) from the line equation into the parabola equation:

• Parabola equation: \(y^2 = 2x\)
• Line equation: \(y = x\)
• Substitute \(y\) in the parabola equation: \((x)^2 = 2x\)
• This simplifies to \(x^2 = 2x\).
• Rearranging the terms gives: \(x^2 - 2x = 0\).
• Factoring out \(x\), we get: \(x(x - 2) = 0\).
• The solutions for \(x\) are \(x = 0\) and \(x = 2\).
• Using the line equation \(y = x\), we find the corresponding \(y\) values:
  • If \(x=0\), then \(y=0\). The first intersection point is \((0, 0)\).
  • If \(x=2\), then \(y=2\). The second intersection point is \((2, 2)\).

The curves intersect at \((0, 0)\) and \((2, 2)\).

Step 2: Set Up the Area Integral

To find the enclosed area, we integrate the difference between the two functions. It is often convenient to express \(x\) in terms of \(y\) when dealing with equations like \(y^2 = 2x\).

• Rewrite the equations with \(x\) as a function of \(y\):
  • Line: \(x = y\)
  • Parabola: \(x = \dfrac{y^2}{2}\)
• The intersection points occur at \(y=0\) and \(y=2\). These will be our limits of integration for \(y\).
• We need to determine which function represents the "rightmost" curve (larger \(x\)) and which represents the "leftmost" curve (smaller \(x\)) within the interval \(0 \le y \le 2\). Let's test a value within this interval, for example, \(y=1\).
  • For the line \(x = y\), when \(y=1\), \(x=1\).
  • For the parabola \(x = \dfrac{y^2}{2}\), when \(y=1\), \(x = \dfrac{1^2}{2} = \dfrac{1}{2}\).
• Since \(1 > \dfrac{1}{2}\) for \(y=1\), the line \(x=y\) is the right curve (\(x_{\text{right}}\)), and the parabola \(x=\dfrac{y^2}{2}\) is the left curve (\(x_{\text{left}}\)) in the interval \([0, 2]\).
• The formula for the area \(A\) is: \(A = \int_{y_{lower}}^{y_{upper}} (x_{\text{right}} - x_{\text{left}}) \, dy\)
• Substituting our functions and limits: \(A = \int_{0}^{2} \left( y - \dfrac{y^2}{2} \right) \, dy\)

Step 3: Evaluate the Integral

Now we compute the definite integral to find the area:

1. Find the antiderivative of the integrand \(\left( y - \dfrac{y^2}{2} \right)\): \(\int \left( y - \dfrac{y^2}{2} \right) \, dy = \dfrac{y^2}{2} - \dfrac{1}{2} \cdot \dfrac{y^3}{3} = \dfrac{y^2}{2} - \dfrac{y^3}{6}\)
2. Evaluate the antiderivative from \(y=0\) to \(y=2\): \(A = \left[ \dfrac{y^2}{2} - \dfrac{y^3}{6} \right]_{0}^{2}\)
3. Substitute the upper and lower limits: \(A = \left( \dfrac{2^2}{2} - \dfrac{2^3}{6} \right) - \left( \dfrac{0^2}{2} - \dfrac{0^3}{6} \right)\)
4. Simplify the expression: \(A = \left( \dfrac{4}{2} - \dfrac{8}{6} \right) - (0 - 0)\) \(A = \left( 2 - \dfrac{4}{3} \right)\)
5. Calculate the final value: \(A = \dfrac{6}{3} - \dfrac{4}{3} = \dfrac{2}{3}\)

Final Area Calculation

The calculated area of the region enclosed between the parabola \(y^2 = 2x\) and the straight line \(y = x\) is \(\dfrac{2}{3}\) square units.