Answer to Question #180265 in Calculus for Prathamesh

Question

Find the area common to the parabolas y2 = 4x and x2 = 4y.

Answer 1

Find the area common to the parabolas $y^2=4x$ and $x^2=4y$ .

Solution:

Points of interception:

$\begin{cases} y=\pm2\sqrt x \\ y=\frac14x^2 \end{cases}$

$\begin{cases} x=0 \\ y=0 \end{cases}$ and $\begin{cases} x=4 \\ y=4 \end{cases}$ .

Area:

$S=\displaystyle\int_0^4 (2\sqrt x-\frac14x^2)dx=$ $(2\cdot \frac23\cdot x^\frac32-\frac14\cdot\frac13\cdot x^3)|_0^4=\frac{2^5}{3}-\frac{4^2}{3}=\frac{16}{3}$ .

Answer: $S=\frac{16}{3}$ .