Question #180265

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


Expert's answer

Find the area common to the parabolas y2=4xy^2=4x and x2=4yx^2=4y .

Solution:



Points of interception:

{y=±2xy=14x2\begin{cases} y=\pm2\sqrt x \\ y=\frac14x^2 \end{cases}

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

Area:

S=∫04(2x−14x2)dx=S=\displaystyle\int_0^4 (2\sqrt x-\frac14x^2)dx= (2⋅23⋅x32−14⋅13⋅x3)∣04=253−423=163(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=163S=\frac{16}{3} .


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