Question

The area enclosed between the curves
$y^2=16xand{x}^2=16y$ is

• A. 2048
• B. 85.33
• C. 2133.33
• D. 1962.67

Answer 1

The correct option is B 85.33

Curve 1 : $y^2=16x$
Curve 2 : $x^2=16y$

Intersection points of curve 1 and 2,

$y^2=16x=16\sqrt{16y}=64\sqrt{y}$
$y^4=64\times 64\times y$
$y^3=64\times 64$
y = 16
and y = 0

then x = 16 and x = 0

Therefore intersection points are P(16, 16) and O(0, 0). The area enclosed between curves 1 and 2 are given by

Area $={\int }_{x_1}^{x_2}{y}_{1}dx-{\int }_{x_2}^{x_1}{y}_{2}dx$

$={\int }_0^{16}\sqrt{16x}dx-{\int }_0^{16}\frac{x^2}{16}dx$

$=4{\left[\frac{x^{3/2}}{3/2}\right]}_0^{16}-{\left[\frac{x^3}{48}\right]}_0^{16}=\frac{8}{3}[64-0]-\frac{1}{48}[{16}^3-0]$
$=170.66-85.33=85.33$