Question

The area bounded by the curves $y^2=8x$and $x^2=8y$ is

• A. $64$ square units
• B. $\frac{64}{3}$ square units
• C. $\frac{8}{3}$ square units
• D. None of these

Answer 1

The correct option is B

$\frac{64}{3}$ square units

Explanation for the correct option:

Find the area bounded by the curves:

Given curves are $y^2=8x$and $x^2=8y$.

Step 1: Find the point of intersection of the curves

$$\begin{gathered} x^2=8y \\ \Rightarrow \frac{x^2}{8}=y \\ \Rightarrow \frac{x^2}{8}=\sqrt{8x}\left[\because y^2=8x\right] \\ \Rightarrow {\left(\frac{x^2}{8}\right)}^2={\left(\sqrt{8x}\right)}^2\left[\text{squaring on both sides}\right] \\ \Rightarrow \frac{x^4}{8^2}=8x \\ \Rightarrow x^4=8^{3}x \\ \Rightarrow x^4-8^{3}x=0 \\ \Rightarrow \left(x^3-8^3\right)x=0 \\ \Rightarrow x=0orx=8 \end{gathered}$$

Step 2: Find the area between the curves

Now the area of required region is given by

$$\begin{gathered} ={\int }_0^8\sqrt{8x}-\frac{x^2}{8}dx \\ =\sqrt{8}{\int }_0^8\sqrt{x}dx-\frac{1}{8}{\int }_0^8{x}^{2}dx \\ =\sqrt{8}{\left[\frac{{\left(x\right)}^{{\displaystyle \frac{3}{2}}}}{\frac{3}{2}}\right]}_0^8-\frac{1}{8}{\left[\frac{x^3}{3}\right]}_0^8 \\ =\frac{2\sqrt{8}}{3}\left[{\left(8\right)}^{\frac{3}{2}}-0\right]-\frac{1}{24}\left[8^3-0\right] \\ =\frac{2\times 8^2}{3}-\frac{8^3}{24} \\ =8^2\left[\frac{2}{3}-\frac{8}{24}\right] \\ =8^2\left[\frac{1}{3}\right] \\ =\frac{64}{3} \end{gathered}$$

Hence, option (B) is the correct answer.