Question

The area of the region bounded by the parabola $x^2=16y$, the lines $y=1$, $y=4$ and the $y$-axis lying in the first quadrant is

• A. $\dfrac{55}{3}$ sq. units
• B. $\dfrac{56}{3}$ sq. units
• C. $\dfrac{52}{3}$ sq. units
• D. $\dfrac{53}{3}$ sq. units

Hint:

For regions bounded by $y$-limits, integrate with respect to $y$ after expressing $x$ in terms of $y$.

Answer 1

The Correct Option is B

Step 1: Expressing $x$ in terms of $y$.
From $x^2=16y$, we have \[ x=4\sqrt{y} \] Since the region lies in the first quadrant and is bounded by the $y$-axis, $x$ varies from $0$ to $4\sqrt{y}$.
Step 2: Setting up the integral.
The area is \[ \int_{y=1}^{4} (4\sqrt{y}-0)\,dy \]
Step 3: Evaluating the integral.
\[ 4\int_{1}^{4} y^{1/2}\,dy =4\left[\frac{2}{3}y^{3/2}\right]_{1}^{4} =\frac{8}{3}(8-1) \]
Step 4: Conclusion.
\[ \text{Area}=\frac{56}{3}\ \text{sq. units} \]