Question

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

• A. $\frac{8}{3}$
• B. $\frac{56}{3}$
• C. $\frac{64}{3}$
• D. $\frac{48}{3}$

Answer 1

The correct option is B $\frac{56}{3}$

Area$\left(A\right)$ of region can be written as
$A={\int }_1^{4}xdy$
$={\int }_1^4\sqrt{16y}dy$
(not taking negative value because area is asked in first quadrant)
$=4{\int }_1^4\sqrt{y}dy$
$=4\times \frac{2}{3}{\left(y^{\frac{3}{2}}\right)}_1^4$
$=\frac{8}{3}(8-1)$
$=\frac{56}{3}$ units