Question

The area of the region described by $\mathrm{A}=\{\left(\mathrm{x},\mathrm{y}\right):{\mathrm{x}}^2+{\mathrm{y}}^2\le 1$ and ${\mathrm{y}}^2\le 1-\mathrm{x}\}$ is

• A. $\left(\frac{\mathrm{\pi }}{2}\right)+\left(\frac{4}{3}\right)$
• B. $\left(\frac{\mathrm{\pi }}{2}\right)-\left(\frac{4}{3}\right)$
• C. $\left(\frac{\mathrm{\pi }}{2}\right)-\left(\frac{2}{3}\right)$
• D. $\left(\frac{\mathrm{\pi }}{2}\right)+\left(\frac{2}{3}\right)$

Answer 1

The correct option is A

$\left(\frac{\mathrm{\pi }}{2}\right)+\left(\frac{4}{3}\right)$

Explanation for the correct answer:

Given,

Step-1 Find the point of intersection of curve:

${\mathrm{x}}^2+{\mathrm{y}}^2\le 1$ and ${\mathrm{y}}^2\le 1-\mathrm{x}$

Plotting the curves, we have:

The shaded area is the required area.

To determine the point of intersection of both curves, we equate the equation,

${\mathrm{x}}^2+{\mathrm{y}}^2=1\dots \left(1\right)$

${\mathrm{y}}^2=1-\mathrm{x}\dots \left(2\right)$

Substitute the value of $y^2$ in equation $\left(1\right)$, we get

$$\begin{gathered} \Rightarrow x^2+1-x=1 \\ \Rightarrow x^2-x=0 \\ \Rightarrow x\left(x-1\right)=0 \\ \Rightarrow x=0\text{and}1 \end{gathered}$$

Substitute the value of $x$in equation $\left(2\right)$.

For , $x=0$

$$\begin{gathered} y^2=1-0 \\ =1 \\ \Rightarrow y=\pm 1 \end{gathered}$$

For, $x=1,y=0$

Thus, the points of intersection are $\left(1,0\right),\left(0,1\right),\left(0,-1\right)$.

Step-2 Area enclosed by the curves:

Area or $A$is given as:

$$\begin{gathered} \mathrm{A}=\text{Area of semi circle}+2{\int }_0^1\sqrt{1-\mathrm{x}}d\mathrm{x} \\ =\frac{{\mathrm{\pi r}}^2}{2}+2{\left[\frac{-{(1-\mathrm{x})}^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{3}{2}}}\right]}_0^1\left(\text{As per formula,}\int x^{n}dx=\frac{x^{n+1}}{n+1}\right) \\ =\frac{\mathrm{\pi }}{2}+2\times \frac{2}{3}\mathbf{}\mathbf{(}\mathit{r}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{\text{for equation}}\mathbf{}{\mathit{x}}^{\mathbf{2}}\mathbf{+}{\mathit{y}}^{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{}{\mathit{r}}^{\mathbf{2}}\mathbf{)} \\ =\frac{\mathrm{\pi }}{2}+\frac{4}{3}\mathrm{sq}\mathrm{units} \end{gathered}$$

Hence, option (A) is the correct answer.