Question Number 72884 by mhmd last updated on 04/Nov/19
$${find}\:{the}\:{area}\:{of}\:{the}\:{region}\:{bounded}\:{by}\:{the}\:{semicircle}\:{y}=\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }\:{and}\:{the}\:{x}=+−{a}\:\:{and}\:{the}\:{line}\:{y}=−{a}\:?\:{by}\:{using}\:{intigiral} \\ $$$${pleas}\:{sir}\:{help}\:{me} \\ $$
Commented by kaivan.ahmadi last updated on 04/Nov/19
![∫_(−a) ^a ((√(a^2 −x^2 ))+a)dx= x=asinθ⇒dx=acosθdθ (√(a^2 −x^2 ))=acosθ { ((x=a⇒θ=(π/2))),((x=−a⇒θ=−(π/2))) :} ∫_(−(π/2)) ^(π/2) a^2 (cosθ+cos^2 θ)dθ=a^2 (sinθ+((θ+(1/2)sin2θ)/2))∣_(−(π/2)) ^(π/2) = a^2 [(1+(π/4))−(−1−(π/4))]=a^2 (2+(π/2))](https://www.tinkutara.com/question/Q72902.png)
$$\int_{−{a}} ^{{a}} \left(\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }+{a}\right){dx}= \\ $$$${x}={asin}\theta\Rightarrow{dx}={acos}\theta{d}\theta \\ $$$$\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }={acos}\theta \\ $$$$\begin{cases}{{x}={a}\Rightarrow\theta=\frac{\pi}{\mathrm{2}}}\\{{x}=−{a}\Rightarrow\theta=−\frac{\pi}{\mathrm{2}}}\end{cases} \\ $$$$\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} {a}^{\mathrm{2}} \left({cos}\theta+{cos}^{\mathrm{2}} \theta\right){d}\theta={a}^{\mathrm{2}} \left({sin}\theta+\frac{\theta+\frac{\mathrm{1}}{\mathrm{2}}{sin}\mathrm{2}\theta}{\mathrm{2}}\right)\mid_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} = \\ $$$${a}^{\mathrm{2}} \left[\left(\mathrm{1}+\frac{\pi}{\mathrm{4}}\right)−\left(−\mathrm{1}−\frac{\pi}{\mathrm{4}}\right)\right]={a}^{\mathrm{2}} \left(\mathrm{2}+\frac{\pi}{\mathrm{2}}\right) \\ $$$$ \\ $$