Question

The area bounded by the curves $y=-\sqrt{-x},x=-\sqrt{-y}$ where $x,y\le 0,$ is

• A. $\frac{1}{3}$
• B. $\frac{1}{4}$
• C. $\frac{1}{5}$
• D. $\frac{1}{2}$

Answer 1

The correct option is A

$\frac{1}{3}$

Explanation for the correct option:

Step 1: Finding Intersection point and draw a diagram

Given curves

$y=-\sqrt{-x}$ and $x=-\sqrt{-y}$

For the intersection of the points

$$\begin{gathered} x=-\sqrt{-\left(-\sqrt{-x}\right)} \\ \Rightarrow x^2=\sqrt{-x} \\ \Rightarrow x^4=-x \\ \Rightarrow x(x^3+1)=0 \\ \Rightarrow x=0,-1 \end{gathered}$$

Step 2: Finding Integral for bounded area

The limit of $x$ lies between $-1$ and $0$ as we move towards the origin

The curve $y=-\sqrt{-x}$ can be written as $y^2=-x$ which is red coloured curve in diagram and

the curve $x=-\sqrt{-y}$ can be written as $x^2=-y$ which is black coloured curve.

The area between the curves will be Red curve minus black curve while taking range of $x$ lies between $-1$ and $0$ so,

area integral A will be

$$\begin{gathered} A={\int }_{-1}^0(-\sqrt{-x})-(-x^2)dx \\ A={\int }_{-1}^0(-\sqrt{-x})+x^2)dx \end{gathered}$$

Step 3: Finding the area integral

Let $$\begin{gathered} -x=t \\ \Rightarrow -dx=dt \end{gathered}$$

and $$\begin{gathered} x^2=t^2 \\ \sqrt{-x}=t \end{gathered}$$

and limits will be $t=1,t=0$

on putting these value in area integral we get,

$$\begin{gathered} A={\int }_1^0\left(\sqrt{t}\right)-t^2)dt \\ A={\int }_0^1(t^2-\sqrt{t}\left)\right)dt \\ A=\left|{\left[\frac{t^3}{3}-\frac{2t^{{\displaystyle \frac{3}{2}}}}{3}\right]}_0^1\right|\mathbf{[}{\mathbf{\int }}_{}^{}{\mathit{x}}^{\mathbf{n}}\mathit{d}\mathit{x}\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{n}\mathbf{+}\mathbf{1}}}{\mathbf{n}\mathbf{+}\mathbf{1}}] \\ A=\frac{1}{3} \end{gathered}$$

Hence, the correct answer is option A.