Question

The area of the region bounded by the curves \(x + 3y^2 = 0\) and \(x + 4y^2 = 1\) is equal to:

• A. \(\frac{1}{3}\)
• B. \(\frac{2}{3}\)
• C. \(\frac{4}{3}\)
• D. \(\frac{5}{3}\)

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The given equations represent two horizontal parabolas. To find the enclosed area, we first find their points of intersection and then integrate the difference between the rightmost curve and the leftmost curve with respect to $y$.
Step 2: Key Formula or Approach:
Area between two curves $x = f(y)$ and $x = g(y)$ from $y = c$ to $y = d$:
$\text{Area} = \int_c^d [f(y) - g(y)] dy$, where $f(y) \ge g(y)$.
Step 3: Detailed Explanation:
The curves are:
1) $x = -3y^2$ (a parabola opening to the left with vertex at the origin).
2) $x = 1 - 4y^2$ (a parabola opening to the left with vertex at $(1, 0)$).
Find the points of intersection by equating the $x$ values:
$-3y^2 = 1 - 4y^2$
$y^2 = 1 \implies y = \pm 1$.
So, the intersection limits are from $y = -1$ to $y = 1$.
Within this interval $y \in [-1, 1]$, we determine which curve is to the right (has larger $x$):
At $y = 0$, $x_1 = 0$ and $x_2 = 1$. So, $x = 1 - 4y^2$ is the right curve $f(y)$, and $x = -3y^2$ is the left curve $g(y)$.
Set up the integral:
$\text{Area} = \int_{-1}^1 [(1 - 4y^2) - (-3y^2)] dy$
$= \int_{-1}^1 (1 - y^2) dy$
Since the integrand is an even function:
$= 2 \int_0^1 (1 - y^2) dy$
$= 2 \left[ y - \frac{y^3}{3} \right]_0^1$
$= 2 \left( 1 - \frac{1}{3} \right) = 2 \left( \frac{2}{3} \right) = \frac{4}{3}$.
Step 4: Final Answer:
The area of the region is $\frac{4}{3}$.