Question

The parabola $y^2=4x$ and $x^2=4y$ divided the square region bounded by the lines $x=4,y=4$ and the coordinate axes. If $S_1,S_2,S_3$ are respectively the areas of these parts numbered from top to bottom, then $S_1:S_2:S_3$ is equal to

• A. $1:1:1$
• B. $2:1:2$
• C. $1:2:3$
• D. $1:2:1$

Answer 1

The correct option is A

$1:1:1$

Step 1: Find the area of $S_1$ and $S_3$

It is clear from the figure that

$$\begin{gathered} S_1=S_3={\int }_0^{4}ydx \\ ={\int }_0^4\frac{x^2}{4}dx \\ =\frac{1}{4}{\left[\frac{x^3}{3}\right]}_0^4 \\ =\frac{1}{12}64 \\ =\frac{16}{3}sq.\mathrm{units} \\ \therefore {\mathrm{S}}_1={\mathrm{S}}_3=\frac{16}{3}sq.\mathrm{units} \end{gathered}$$

Step 2: Find the area of $S_2$

Now,

$$\begin{gathered} S_2+S_3={\int }_0^4\sqrt{4x}dx \\ \Rightarrow S_2+S_3=2{\left[\frac{x^{{\displaystyle \frac{3}{2}}}}{{}^{\frac{3}{2}}}\right]}_0^4 \\ \Rightarrow S_2+S_3=\frac{4}{3}\times 8 \\ \Rightarrow S_2+S_3=\frac{32}{3} \\ \Rightarrow S_2=\frac{32}{3}-S_3 \\ \Rightarrow S_2=\frac{32}{3}-\frac{16}{3} \\ \Rightarrow S_2=\frac{16}{3}sq.\mathrm{units} \end{gathered}$$

Therefore $S_1:S_2:S_3=\frac{16}{3}:\frac{16}{3}:\frac{16}{3}=1:1:1$

Hence, the correct option is $\left(\mathrm{A}\right)$.