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

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

Answers (1)

\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}

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