Question-58447

Question Number 58447 by Tawa1 last updated on 23/Apr/19

Commented by Tawa1 last updated on 23/Apr/19

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{Area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{Red}\:\mathrm{portion} \\ $$

Answered by MJS last updated on 23/Apr/19

halfcircle: y=4−(√(8x−x^2 )) line: y=(x/2) intersection = (((8/5)),((4/5)) ) red area ∫_0 ^(8/5) (x/2)dx+∫_((8 )/5) ^4 (4−(√(8x−x^2 )))dx=((32)/5)−4π+16arctan (1/2) ≈1.25199

$$\mathrm{halfcircle}:\:{y}=\mathrm{4}−\sqrt{\mathrm{8}{x}−{x}^{\mathrm{2}} } \\ $$$$\mathrm{line}:\:{y}=\frac{{x}}{\mathrm{2}} \\ $$$$\mathrm{intersection}\:=\:\begin{pmatrix}{\frac{\mathrm{8}}{\mathrm{5}}}\\{\frac{\mathrm{4}}{\mathrm{5}}}\end{pmatrix} \\ $$$$\mathrm{red}\:\mathrm{area} \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\mathrm{8}}{\mathrm{5}}} {\int}}\frac{{x}}{\mathrm{2}}{dx}+\underset{\frac{\mathrm{8}\:\:}{\mathrm{5}}} {\overset{\mathrm{4}} {\int}}\left(\mathrm{4}−\sqrt{\mathrm{8}{x}−{x}^{\mathrm{2}} }\right){dx}=\frac{\mathrm{32}}{\mathrm{5}}−\mathrm{4}\pi+\mathrm{16arctan}\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\approx\mathrm{1}.\mathrm{25199} \\ $$

Commented by Tawa1 last updated on 23/Apr/19

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by Tawa1 last updated on 23/Apr/19

Sir, how is half circle y = 4 − (√(8x − x^2 ))

$$\mathrm{Sir},\:\mathrm{how}\:\mathrm{is}\:\mathrm{half}\:\mathrm{circle}\:\:\:\mathrm{y}\:=\:\mathrm{4}\:−\:\sqrt{\mathrm{8x}\:−\:\mathrm{x}^{\mathrm{2}} } \\ $$

Commented by MJS last updated on 23/Apr/19

I put the left handed lower bottom verticle of the rectangle A= ((0),(0) ) ⇒ center of circle = = ((4),(4) ) ⇒ circle: (x−4)^2 +(y−4)^2 =4^2 ⇒ ⇒ y=4±(√(8x−x^2 ))

$$\mathrm{I}\:\mathrm{put}\:\mathrm{the}\:\mathrm{left}\:\mathrm{handed}\:\mathrm{lower}\:\mathrm{bottom}\:\mathrm{verticle} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{rectangle}\:{A}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:\Rightarrow\:\mathrm{center}\:\mathrm{of}\:\mathrm{circle}\:= \\ $$$$=\begin{pmatrix}{\mathrm{4}}\\{\mathrm{4}}\end{pmatrix}\:\Rightarrow\:\mathrm{circle}:\:\left({x}−\mathrm{4}\right)^{\mathrm{2}} +\left({y}−\mathrm{4}\right)^{\mathrm{2}} =\mathrm{4}^{\mathrm{2}} \:\Rightarrow \\ $$$$\Rightarrow\:{y}=\mathrm{4}\pm\sqrt{\mathrm{8}{x}−{x}^{\mathrm{2}} } \\ $$

Commented by Tawa1 last updated on 23/Apr/19

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Answered by mr W last updated on 23/Apr/19

Commented by mr W last updated on 23/Apr/19

α=tan^(−1) (4/8)=tan^(−1) (1/2) θ=((π−2α)/2)=(π/2)−α A_(red shade) =4×4−((π4^2 )/4)=16−4π A_(green shade) =(4^2 /2)(2θ−sin 2θ)=16(θ−sin θ cos θ) =16((π/2)−tan^(−1) (1/2)−(2/( (√5)))×(1/( (√5))))=16((π/2)−(2/5)−tan^(−1) (1/2)) A_(red) =((4×8)/2)−16((π/2)−(2/5)−tan^(−1) (1/2))−(16−4π) =((32)/5)+16 tan^(−1) (1/2)−4π =1.25199

$$\alpha=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{4}}{\mathrm{8}}=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\theta=\frac{\pi−\mathrm{2}\alpha}{\mathrm{2}}=\frac{\pi}{\mathrm{2}}−\alpha \\ $$$${A}_{{red}\:{shade}} =\mathrm{4}×\mathrm{4}−\frac{\pi\mathrm{4}^{\mathrm{2}} }{\mathrm{4}}=\mathrm{16}−\mathrm{4}\pi \\ $$$${A}_{{green}\:{shade}} =\frac{\mathrm{4}^{\mathrm{2}} }{\mathrm{2}}\left(\mathrm{2}\theta−\mathrm{sin}\:\mathrm{2}\theta\right)=\mathrm{16}\left(\theta−\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta\right) \\ $$$$=\mathrm{16}\left(\frac{\pi}{\mathrm{2}}−\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\right)=\mathrm{16}\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{2}}{\mathrm{5}}−\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$ \\ $$$${A}_{{red}} =\frac{\mathrm{4}×\mathrm{8}}{\mathrm{2}}−\mathrm{16}\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{2}}{\mathrm{5}}−\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}\right)−\left(\mathrm{16}−\mathrm{4}\pi\right) \\ $$$$=\frac{\mathrm{32}}{\mathrm{5}}+\mathrm{16}\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}−\mathrm{4}\pi \\ $$$$=\mathrm{1}.\mathrm{25199} \\ $$

Commented by Tawa1 last updated on 23/Apr/19

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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