Question

Find thrice the area of the region bounded by the curves $x^2=16y,y=1,y=4$ and the $y$-axis lying in the first quadrant.

Answer 1

$x^2=16y$

$y=\frac{x^2}{16}$

$area={\int }_1^4(y{)}^{\frac{1}{2}}=2\times \frac{4(y{)}^{\frac{3}{2}}}{3}{|}_1^4$

$=\frac{8}{3}[8-1]=\frac{56}{3}$

thrice the area of the region $=56$