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Find-0-4-16-x-2-dx-




Question Number 207170 by hardmath last updated on 08/May/24
Find:   ∫_0 ^( 4)  (√(16 − x^2 )) dx = ?
$$\mathrm{Find}:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{4}} \:\sqrt{\mathrm{16}\:−\:\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx}\:=\:? \\ $$
Commented by MM42 last updated on 08/May/24
x=4sinθ⇒dx=4cosθdθ  ∫_0 ^(π/2)  16cos^2 θdθ=∫_0 ^(π/2) 8(1+cos2θ)dθ  =8(θ+(1/2)sin2θ)]_0 ^(π/2)   =4π  ✓
$${x}=\mathrm{4}{sin}\theta\Rightarrow{dx}=\mathrm{4}{cos}\theta{d}\theta \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{16}{cos}^{\mathrm{2}} \theta{d}\theta=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{8}\left(\mathrm{1}+{cos}\mathrm{2}\theta\right){d}\theta \\ $$$$\left.=\mathrm{8}\left(\theta+\frac{\mathrm{1}}{\mathrm{2}}{sin}\mathrm{2}\theta\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$=\mathrm{4}\pi\:\:\checkmark \\ $$$$ \\ $$
Commented by hardmath last updated on 08/May/24
Answer:  4π
$$\mathrm{Answer}:\:\:\mathrm{4}\pi \\ $$
Commented by MM42 last updated on 08/May/24
★  ∫_0 ^a (√(a^2 −x^2  ))=^(x=asinθ) ∫_0 ^(π/2) a^2 cos^2 θdθ  =(((πa^2 )/4))  ✓
$$\bigstar \\ $$$$\int_{\mathrm{0}} ^{{a}} \sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} \:}\overset{{x}={asin}\theta} {=}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {a}^{\mathrm{2}} {cos}^{\mathrm{2}} \theta{d}\theta \\ $$$$=\left(\frac{\pi{a}^{\mathrm{2}} }{\mathrm{4}}\right)\:\:\checkmark \\ $$$$ \\ $$
Commented by hardmath last updated on 10/May/24
thank you dear professor
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor} \\ $$
Answered by Skabetix last updated on 09/May/24
=∫_0 ^4 (√(4^2 −x^2 ))  Or ∫_0 ^a (√(a^2 −x^2  ))dx=(a^2 /4)π  So here ∫_0 ^4 (√(4^2 −x^2 ))dx=(4^2 /4)π=((16)/4)π=4π
$$=\int_{\mathrm{0}} ^{\mathrm{4}} \sqrt{\mathrm{4}^{\mathrm{2}} −{x}^{\mathrm{2}} } \\ $$$${Or}\:\int_{\mathrm{0}} ^{{a}} \sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} \:}{dx}=\frac{{a}^{\mathrm{2}} }{\mathrm{4}}\pi \\ $$$${So}\:{here}\:\int_{\mathrm{0}} ^{\mathrm{4}} \sqrt{\mathrm{4}^{\mathrm{2}} −{x}^{\mathrm{2}} }{dx}=\frac{\mathrm{4}^{\mathrm{2}} }{\mathrm{4}}\pi=\frac{\mathrm{16}}{\mathrm{4}}\pi=\mathrm{4}\pi \\ $$

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