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  • Sketch the region and x-axis. Find the area of the region using integration.

Sketch the region  and x-axis. Find the area of the region using integration.

Answers (1)

\begin{aligned} &y=\sqrt{4-x^{2}}\\ &\text { Square both sides }\\ &\Rightarrow y^{2}=4-x^{2}\\ &\Rightarrow x^{2}+y^{2}=4\\ &\Rightarrow x^{2}+y^{2}=2^{2} \end{aligned} \\

The above equation depicts a circle with origin and radius 2 

Now in  \mathrm{y}=\sqrt{4-\mathrm{x}^{2}} \ \ -2 \leq \mathrm{x} \leq 2 \ and \ \mathrm{y} \geq 0\text{ which means }\mathrm{x} \: \ and \: \ \mathrm{y} \text{ both positive or } \mathrm{x} \text{ negative and } \mathrm{y} \text{ positive hence the curve }\mathrm{y}=\sqrt{4-\mathrm{x}^{2}} \text{has to be above \mathrm{X} -axis }in\ 1^{\mathrm{st}} \: \: and \: \: 2^{\text {nd }} quadrant\text{Hence the graph of y}=\sqrt{4-x^2} \text{will be graph of circle } \mathrm{x}^{2}+\mathrm{y}^{2}=2^{2} \text{lying only above X -axis }

Equation of X axis is y = 0

For point of interaction, 

Y = 0
\\ \item X\textsuperscript{2} = 4 \\ \item X = \pm 2 \\

Point of interaction are (-2, 0) and (2, 0) 

The below figure shows the area 

Now let us find the area

 y=\sqrt{4-x^{2}}

Integrate from -2 to 2 

\Rightarrow \int_{-2}^{2} \mathrm{ydx}=\int_{-2}^{2} \sqrt{4-\mathrm{x}^{2}} \mathrm{~d} \mathrm{x} \\

\\\begin{aligned} &\text { Using uv rule of integration where u and } v \text { are functions of } x\\ &\int_{a}^{b} u v d x=\left[u \int \operatorname{vdx}\right]_{a}^{b}-\int_{a}^{b}\left(u^{\prime} \int v d x\right) d x\\ &\text { Here } u=\sqrt{4-x^{2}} \text { and } v=1\\ &\text { Hence } u^{\prime}=\frac{1}{2}\left(4-x^{2}\right)^{\frac{1}{2}-1}(-2 x)=\frac{-x}{\sqrt{4-x^{2}}}\\ &\Rightarrow \int_{-2}^{2} y d x=\left[\sqrt{4-x^{2}} \int 1 d x\right]_{-2}^{2}-\int_{-2}^{2}\left(\frac{-x}{\sqrt{4-x^{2}}}\right)(1 d x) d x\\ &\Rightarrow \int_{-2}^{2} y d x=\left[\sqrt{4-x^{2}}(x)\right]_{-2}^{2}-\int_{-2}^{2}\left(\frac{-x^{2}}{\sqrt{4-x^{2}}}\right) d x\\ &\Rightarrow \int_{-2}^{2} \mathrm{ydx}=\left(\sqrt{4-2^{2}}(2)\right)-\left(\sqrt{4-(-2)^{2}}(-2)\right)-\int_{-2}^{2}\left(\frac{4-\mathrm{x}^{2}-4}{\sqrt{4-\mathrm{x}^{2}}}\right) \mathrm{d} \mathrm{x} \end{aligned}

\\ \Rightarrow \int_{-2}^{2} y d x=-\int_{-2}^{2}\left(\frac{4-x^{2}}{\sqrt{4-x^{2}}}-\frac{4}{\sqrt{4-x^{2}}}\right) d x \\ \Rightarrow \int_{-2}^{2} y d x=-\int_{-2}^{2} \sqrt{4-x^{2}} d x+\int_{-2}^{2} \frac{4}{\sqrt{4-x^{2}}} d x \\ \text { But } y=\sqrt{4-x^{2}} \\ \Rightarrow \int_{-2}^{2} y d x=-\int_{-2}^{2} y d x+\int_{-2}^{2} \frac{4}{\sqrt{4-x^{2}}} d x \\ \Rightarrow \int_{-2}^{2} y d x+\int_{-2}^{2} y d x=\int_{-2}^{2} \frac{4}{\sqrt{4-x^{2}}} d x \\ \Rightarrow \int_{-2}^{2} y d x=2 \int_{-2}^{2} \frac{1}{\sqrt{2^{2}-x^{2}}} d x

\\\begin{aligned} &\text { We know that } \int \frac{1}{\sqrt{a^{2}-x^{2}}} d x=\sin ^{-1} \frac{x}{a}\\ &\Rightarrow \int_{-2}^{2} \mathrm{y} \mathrm{dx}=2\left[\sin ^{-1} \frac{\mathrm{x}}{2}\right]_{-2}^{2}\\ &\Rightarrow \int_{-2}^{2} \mathrm{y} \mathrm{dx}=2\left(\sin ^{-1} \frac{2}{2}-\sin ^{-1} \frac{-2}{2}\right)\\ &\Rightarrow \int_{-2}^{2} \mathrm{y} \mathrm{dx}=2\left(\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)\right)\\ &\Rightarrow \int_{-2}^{2} \mathrm{y} \mathrm{dx}=2\left(\frac{\pi}{2}+\frac{\pi}{2}\right)\\ &\Rightarrow \int_{-2}^{2} \mathrm{y} \mathrm{dx}=2 \pi\\ &\text { Hence area is } 2 \pi \text { unit }^{2} \end{aligned}

 

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