Question

Find the area bounded by the curve $y=x^2$ and the line $y=4$.

Answer 1

Given that $y=4$

Let the line $AB$ represents $y=4$

Also,$y=x^2,x^2=y$

Let $AOB$ represent $x^2=y$

We have to find the area of $AOBA$

Area of $AOBA=2\times$ Area $BONB=2{\int }_0^{4}xdy$

We know that $x^2=y$

$\Rightarrow x=\pm \sqrt{y}$

Since $BONB$ is in first quadrant.

We use $x=+\sqrt{y}$

Area of $AOBA=2{\int }_0^{4}xdy$

$=2{\int }_0^4\sqrt{y}dy$

$=2{\left[\frac{y^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]}_0^4$

$=\frac{4}{3}{\left[y^{\frac{3}{2}}\right]}_0^4$

$=\frac{4}{3}[4^{\frac{3}{2}}-0]$

$=\frac{4}{3}\times 8=\frac{32}{3}$sq.units.