Question

The area (in sq. units) of the region bounded by the parabola $x = \frac{y^2}{2}$ and the line $x = y + 4$ is

• A. 30
• B. $\frac{53}{3}$
• C. 16
• D. 18

Hint:

Logic Tip: When a parabola opens horizontally (i.e., its axis of symmetry is parallel to the x-axis), it is almost always more efficient to integrate with respect to $y$. Integrating with respect to $x$ would require splitting the area into multiple integrals and dealing with square root functions.

Answer 1

The Correct Option is D

Concept:
The area bounded by two curves $x = f(y)$ and $x = g(y)$ between the horizontal lines $y = c$ and $y = d$ is given by the integral: $$\text{Area} = \int_{c}^{d} [\text{Right Curve } (x_R) - \text{Left Curve } (x_L)] \, dy$$ Integrating with respect to $y$ is much easier here because the curves are explicitly defined as $x$ in terms of $y$.
Step 1: Find the points of intersection of the two curves.
Equate the two expressions for $x$ to find the intersection points on the y-axis: $$x = \frac{y^2}{2} \quad \text{and} \quad x = y + 4$$ $$\frac{y^2}{2} = y + 4$$ Multiply by 2 to clear the fraction: $$y^2 = 2y + 8$$ $$y^2 - 2y - 8 = 0$$ Factor the quadratic equation: $$(y - 4)(y + 2) = 0$$ So, the limits of integration are $y = -2$ to $y = 4$.
Step 2: Set up the definite integral for the area.
To determine which curve is on the "right" ($x_R$) and which is on the "left" ($x_L$), test a point in the interval $(-2, 4)$, such as $y=0$: For $x = y + 4$, $x = 0 + 4 = 4$. For $x = y^2/2$, $x = 0^2/2 = 0$. Since $4>0$, the line is the right curve, and the parabola is the left curve. $$\text{Area} = \int_{-2}^{4} \left( (y + 4) - \left(\frac{y^2}{2}\right) \right) dy$$
Step 3: Evaluate the integral.
Integrate the polynomial with respect to $y$: $$\text{Area} = \left[ \frac{y^2}{2} + 4y - \frac{y^3}{6} \right]_{-2}^{4}$$ Substitute the upper limit ($y=4$): $$\text{Upper} = \frac{4^2}{2} + 4(4) - \frac{4^3}{6} = \frac{16}{2} + 16 - \frac{64}{6} = 8 + 16 - \frac{32}{3} = 24 - \frac{32}{3} = \frac{40}{3}$$ Substitute the lower limit ($y=-2$): $$\text{Lower} = \frac{(-2)^2}{2} + 4(-2) - \frac{(-2)^3}{6} = \frac{4}{2} - 8 - \frac{-8}{6} = 2 - 8 + \frac{4}{3} = -6 + \frac{4}{3} = \frac{-14}{3}$$ Subtract the lower bound evaluation from the upper bound evaluation: $$\text{Area} = \text{Upper} - \text{Lower} = \frac{40}{3} - \left(\frac{-14}{3}\right)$$ $$\text{Area} = \frac{40}{3} + \frac{14}{3} = \frac{54}{3} = 18$$