MHT CET · Maths · Area Under Curves
The area of the region bounded by the parabola \(x^{2}=16 y, y=1, y=4\) and the
Y-axis lying in the first quartrant is
- A \(\frac{55}{3}\) sq. units
- B \(\frac{56}{3}\) sq. units
- C \(\frac{52}{3}\) sq. units
- D \(\frac{53}{3}\) sq. units
Answer & Solution
Correct Answer
(B) \(\frac{56}{3}\) sq. units
Step-by-step Solution
Detailed explanation
(A)
The equation of curve is \(x^{2}=16 y \Rightarrow x=4 \sqrt{y}\) Required area is shaded.
\(\begin{aligned}
A &=\int_{1}^{4} 4 \sqrt{y} d y=4 \int_{1}^{4} y^{\frac{1}{2}} d y=4\left[\frac{y^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}\right]_{1}^{4} \\
&=\frac{8}{3}(4 \sqrt{4}-1)=\frac{8 \times 7}{3}=\frac{56}{3}
\end{aligned}\)

The equation of curve is \(x^{2}=16 y \Rightarrow x=4 \sqrt{y}\) Required area is shaded.
\(\begin{aligned}
A &=\int_{1}^{4} 4 \sqrt{y} d y=4 \int_{1}^{4} y^{\frac{1}{2}} d y=4\left[\frac{y^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}\right]_{1}^{4} \\
&=\frac{8}{3}(4 \sqrt{4}-1)=\frac{8 \times 7}{3}=\frac{56}{3}
\end{aligned}\)

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