#### Determine the area under the curve $y=\sqrt{a^2-x^2}$  included between the lines x = 0 and x = a

\begin{aligned} &y=\sqrt{a^{2}-x^{2}}\\ &\text { Squaring both sides }\\ &\Rightarrow y^{2}=a^{2}-x^{2}\\ &\Rightarrow x^{2}+y^{2}=a^{2} \end{aligned} \\

The above equation is of a circle having centre as (0, 0) and radius a

Now in  $\mathrm{y}=\sqrt{\mathrm{a}^{2}-\mathrm{x}^{2}} -a \leq \mathrm{x} \leq \mathrm{a}$  and  $\mathrm{y} \geq 0$  which means  x and  y  both positive or x  negative and  y  positive hence the curve  $y=\sqrt{a^{2}-x^{2}}$  has to be above  X  -axis in  $1^{s t}$  and  $2^{\text {nd }}$  quadrant

x=0  is equation of  Y  -axis and x=a  is a line parallel to Y-axis passing through  (a, 0)

So we have to integrate $y=\sqrt{a^2-x^2}$  from 0 to a

So we have to integrate $y=\sqrt{a^2-x^2}$ from 0 to a

let us find area under curve

$\\ y=\sqrt{a^{2}-x^{2}} \\$

Integrate from 0 to a

$\Rightarrow \int_{0}^{a} \mathrm{ydx}=\int_{0}^{a} \sqrt{\mathrm{a}^{2}-\mathrm{x}^{2}} \mathrm{dx}$

Using uv rule of integration where u and  v  are functions of

$\int_{a}^{b} u v d x=\left[u \int v d x\right]_{a}^{b}-\int_{a}^{b}\left(u^{\prime} \int v d x\right) d x$

Here  $u=\sqrt{a^{2}-x^{2}}$  and  v=1

Hence

$\mathrm{u}^{\prime}=\frac{1}{2}\left(\mathrm{a}^{2}-\mathrm{x}^{2}\right)^{\frac{1}{2}-1}(-2 \mathrm{x})=\frac{-\mathrm{x}}{\sqrt{\mathrm{a}^{2}-\mathrm{x}^{2}}}$

$\\ \begin{array}{l} \Rightarrow \int_{0}^{a} y d x=\left[\sqrt{a^{2}-x^{2}} \int_{0}^{a} d x\right]_{0}^{a}-\int_{0}^{a}\left(\frac{-x}{\sqrt{a^{2}-x^{2}}} \int_{0}^{a} d x\right) d x \\ \Rightarrow \int_{0}^{a} y d x=\left[\sqrt{a^{2}-x^{2}}(x)\right]_{0}^{a}-\int_{0}^{a}\left(\frac{-x^{2}}{\sqrt{a^{2}-x^{2}}}\right) \mathrm{d} x \\ \Rightarrow \int_{0}^{a} y d x=\left(\sqrt{a^{2}-a^{2}}(a)\right)-\left(\sqrt{a^{2}-0^{2}}(0)\right)-\int_{0}^{a}\left(\frac{a^{2}-x^{2}-a^{2}}{\sqrt{a^{2}-x^{2}}}\right) \mathrm{d} x \\ \Rightarrow \int_{0}^{a} y d x=-\int_{0}^{a}\left(\frac{a^{2}-x^{2}}{\sqrt{a^{2}-x^{2}}}-\frac{a^{2}}{\sqrt{a^{2}-x^{2}}}\right) d x \\ \end{array}$

$\\ \begin{array}{l} \Rightarrow \int_{0}^{a} y d x=-\int_{0}^{a} \sqrt{a^{2}-x^{2}} d x+\int_{0}^{a} \frac{a^{2}}{\sqrt{a^{2}-x^{2}}} d x \\ \text { But } y=\sqrt{a^{2}-x^{2}} \\ \Rightarrow \int_{0}^{a} y d x=-\int_{0}^{a} y d x+\int_{0}^{a} \frac{a^{2}}{\sqrt{a^{2}-x^{2}}} d x \end{array}$
\begin{aligned} &\Rightarrow \int_{0}^{a} y d x+\int_{0}^{a} y d x=\int_{0}^{a} \frac{a^{2}}{\sqrt{a^{2}-x^{2}}} d x\\ &\Rightarrow 2 \int_{0}^{a} y d x=a^{2} \int_{0}^{a} \frac{1}{\sqrt{a^{2}-x^{2}}} d x\\ &\Rightarrow \int_{0}^{a} y d x=\frac{a^{2}}{2} \int_{0}^{a} \frac{1}{\sqrt{a^{2}-x^{2}}} d x\\ \end{aligned} \\

$\text{We know that }$  $\int_{0}^{a} \frac{1}{\sqrt{a^{2}-x^{2}}} d x = \sin ^{-1} \frac{x}{a}$

$\\ \Rightarrow \int_{0}^{a} y d x=\frac{a^{2}}{2}\left[\sin ^{-1} \frac{x}{a}\right]_{0}^{a}\\ \Rightarrow \int_{0}^{a} y d x=\frac{a^{2}}{2}\left(\sin ^{-1} \frac{a}{a}-\sin ^{-1} \frac{0}{2}\right) \\$
$\begin{array}{l} \Rightarrow \int_{0}^{a} y d x=\frac{a^{2}}{2}\left(\frac{\pi}{2}-0\right) \\ \Rightarrow \int_{0}^{a} y d x=\frac{\pi a^{2}}{4} \\ \text { Hence area bounded }=\frac{\pi a^{2}}{4} \text { unit }^{2} \end{array} \\ \vspace{\baselineskip} \vspace{\baselineskip} \vspace{\baselineskip} \vspace{\baselineskip} \vspace{\baselineskip} \vspace{\baselineskip}$