Question

The area of the region enclosed by
\(y≤4x^2, x2≤9y\ and\ y≤4,\)
is equal to

• A. \(\frac{40}{3}\)
• B. \(\frac{56}{3}\)
• C. \(\frac{112}{3}\)
• D. \(\frac{80}{3}\)

Answer 1

The Correct Option is D

To find the area of the region enclosed by the curves \( y \leq 4x^2 \), \( x^2 \leq 9y \), and \( y \leq 4 \), we follow these steps:

1. First, determine the boundary curves:
   1. The equation \( y = 4x^2 \) is a parabola opening upwards.
   2. The equation \( x^2 = 9y \) can be rewritten as \( y = \frac{x^2}{9} \), which is also a parabola, but opening upwards and stretched vertically.
   3. The line \( y = 4 \) is a horizontal line.
2. Identify the points of intersection of these boundary curves:
   1. Find where \( y = 4x^2 \) intersects \( y = \frac{x^2}{9} \): \[ 4x^2 = \frac{x^2}{9} \] \[ 36x^2 = x^2 \] \[ x^2 (36 - 1) = 0 \rightarrow x^2 = 0 \rightarrow x = 0 \]
   2. Intersection of \( y = 4x^2 \) and \( y = 4 \): \[ 4x^2 = 4 \rightarrow x^2 = 1 \rightarrow x = \pm1 \]
   3. Intersection of \( x^2 = 9y \) and \( y = 4 \): \[ x^2 = 9 \cdot 4 \rightarrow x^2 = 36 \rightarrow x = \pm 6 \]
3. Visualize the area bounded by these curves:
4. Calculate the area:
   1. Set up the integral for \( y = 4x^2 \) to \( y = 4 \) between \( x = -1 \) and \( x = 1 \). \[ \text{Area} = 2 \int_0^1 (4 - 4x^2) \, dx \]
   2. Compute the definite integral: \[ \text{Area} = 2 \left[ 4x - \frac{4x^3}{3} \right]_0^1 = 2 \left( 4\cdot1 - \frac{4\cdot1^3}{3} \right) \] \[ = 2 \times \left(4 - \frac{4}{3}\right) = 2 \times \frac{8}{3} = \frac{16}{3} \]
   3. Calculate the area for the portion \( x^2 = 9y \) to \( y = 4 \) from \( x = 1 \) to \( x = 6 \): \[ \text{Area} = 2 \int_1^6 \left( 4 - \frac{x^2}{9} \right) \, dx \]
   4. Calculate this area: \[ \text{Area} = 2 \left[ 4x - \frac{x^3}{27} \right]_1^6 = 2 \left( 24 - \frac{216}{27} \right) \] \[ = 2 \times \left( 24 - 8 \right) = 2 \times 16 = 32 \]
5. Add both computed areas: \[ \text{Total Area} = \frac{16}{3} + \frac{64}{3} = \frac{80}{3} \]

Hence, the area of the region enclosed by the given curves is \(\frac{80}{3}\).