Question

The area (in sq units) bounded by the curves $y^2=4x$ and $x^2=4y$ in the plane is

• A. $\frac{8}{3}$
• B. $\frac{16}{3}$
• C. $\frac{32}{3}$
• D. $\frac{64}{3}$

Answer 1

The correct option is B

$\frac{16}{3}$

Explanation for correct options:

Determining the area bounded by the curves:

Given that the curves are $y^2=4x$ and $x^2=4y$

$y^2=4x\to \left(i\right)$ is a rightward parabola.

$x^2=4y$

$\therefore y=\frac{x^2}{4}\to \left(ii\right)$ is an upward parabola.

Now the graph of the given curves is as shown:

Then the point of intersection of the two curves is given by substituting the equation $\left(ii\right)$ in equation $\left(i\right)$.

$$\begin{gathered} y^2=4x \\ \Rightarrow {\left(\frac{x^2}{4}\right)}^2=4x \\ \Rightarrow \frac{x^4}{16}=4x \\ \Rightarrow x^4=4\times 16x \\ \Rightarrow x^4-64x=0 \\ \Rightarrow x\left(x^3-64\right)=0 \\ \Rightarrow x=0\text{or}x=4 \end{gathered}$$

Then the given curve is rewritten as

$$\begin{gathered} y^2=4x \\ \Rightarrow y=\sqrt{4x} \\ =2\sqrt{x} \end{gathered}$$

$$\begin{gathered} x^2=4y \\ \Rightarrow y=\frac{x^2}{4} \end{gathered}$$

Now the area of the bounded region is given by

$\begin{array}{l}A={\int }_0^4\left(2\sqrt{x}-\frac{x^2}{4}\right)dx\\ ={\left[\left(2\times \frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right)-\frac{x^3}{12}\right]}_0^4\\ ={\left[\left(\frac{4}{3}\times x^{\frac{3}{2}}\right)-\frac{x^3}{12}\right]}_0^4\\ =\left[\left(\left(\frac{4}{3}\times 4^{\frac{3}{2}}\right)-\frac{4^3}{12}\right)-0\right]\\ =\left[\frac{4\times 8}{3}-\frac{64}{12}\right]\\ =\left[\frac{32}{3}-\frac{64}{12}\right]\\ =\left[\frac{128-64}{12}\right]\\ =\frac{64}{12}\\ =\frac{16}{3}\end{array}$

Therefore the area bounded by the given curve is equal to $\frac{16}{3}$square units

Hence, option (B) is the correct answer