Question

The area bounded by the two parabolas $y^2=8x$ and $x^2=8y$ is

• A. $64$ sq. units
• B. $\frac{64}{3}$ sq, units
• C. $\frac{32}{3}$ sq. units
• D. $\frac{1}{3}$ sq. units

Answer 1

Answer: (B)

$\frac{64}{3}$ sq, units

$y^2=8x$ and $x^2=8y$
the points of intersection
$x^2/8=\sqrt{8x}$
$\frac{x^3}{8^2}=8$
x=0 and x=8
So, area b//w curves is
$\stackrel{8}{\underset{0}{\int }}(\sqrt{8x}-x^2/8)dx$
$\sqrt{8}\stackrel{8}{\underset{0}{\int }}\sqrt{x}dx-\stackrel{8}{\underset{0}{\int }}\left(\frac{x^3}{8}\right)dx=\frac{2\sqrt{8}}{3}(x{)}^{3/2}{|}_0^8-1/8\frac{x^3}{3}{|}_0^8$
$2/38^2-1/8\left(8^3/3\right)$
$=\frac{{2.8}^2}{3}-8^2/3=\frac{64}{3}squnits$