Application of the Integrals
86805
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
1 \(\frac{56}{3}\) sq.units
2 \(\frac{52}{3}\) sq.units
3 \(\frac{53}{3}\) sq.units
4 \(\frac{55}{3}\) sq.units
Explanation:
(A) : Given parabola, \(\mathrm{x}^{2}=16 \mathrm{y}\) \(\mathrm{x}=4 \sqrt{\mathrm{y}}\)
And, \(\mathrm{y}=1, \mathrm{y}=4\)
Required area, \(\mathrm{A}=\int_{1}^{4} \mathrm{x} . \mathrm{dy}\)
\(A=\int_{1}^{4} 4 \sqrt{y} . d y\)
\(\mathrm{A}=4\left[\frac{\mathrm{y}^{\frac{3}{2}}}{\frac{3}{2}}\right]_{1}^{4}=\frac{4 \times 2}{3}\left[(4)^{3 / 2}-(1)^{3 / 2}\right]=\frac{8}{3}[8-1]\)
\(A=\frac{56}{3}\) sq. units.
