Question

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

Answer 1

Area (A) of region can be written as,
$a={\int }_1^{4}xdy$
$={\int }_1^4\sqrt{16y}\text{dy}$
$=4{\int }_1^4\sqrt{y}\text{dy}$
$=4\times \frac{2}{3}{\left(y^{\frac{3}{2}}\right)}^4$
$=\frac{8}{3}(8-1)$
$=\frac{56}{3}$