A B C D {\displaystyle ABCD} kvadrat stranice 10.
Lik je određen sa
A P D ∧ C P D {\displaystyle APD \land CPD } polukružnice i A D Q B {\displaystyle ADQB } četvrtina kružnice Slika 1
I način
Slika 1
b 2 2 ( π − 2 θ ) + a 2 2 ( 2 θ ) = ( a 2 − b 2 ) ( tan − 1 ( b a ) ) + π b 2 2 − a b {\displaystyle \frac{b^2}{2}(\pi -2\theta) +\frac{a^2}{2}(2\theta) =(a^2-b^2) (\tan^{-1}(\frac{b}{a})) +\frac{\pi b^2}{2} -ab } ( tan − 1 ( b a ) {\displaystyle \tan^{-1} (\frac{b}{a}) } )
Slika 2
( 10 2 − 5 2 ) ( tan − 1 ( 5 10 ) π ∗ 5 2 2 − 10 ∗ 5 = {\displaystyle (10^2-5^2)(\tan^{-1} (\frac{5}{10})\frac{\pi* 5^2}{2}-10*5 = }
75 ∗ tan − 1 ( 0 , 5 ) + 12 , 5 π − 50 {\displaystyle 75* \tan^{-1}(0,5)+12,5 \pi -50 } ( Slika 2)
Slika 3
( 5 2 − 5 2 ) ( tan − 1 ( 5 5 ) π ∗ 5 2 2 − 5 ∗ 5 = 12 , 5 π − 25 {\displaystyle (5^2-5^2)(\tan^{-1} (\frac{5}{5})\frac{\pi* 5^2}{2}-5*5 = 12,5\pi - 25 } (Slika 3)
75 ∗ tan − 1 ( 0 , 5 ) + 12 , 5 π − 50 − 12 , 5 π − 25 = 75 ∗ tan − 1 ( 0 , 5 ) − 25 ≈ 9 , 77 {\displaystyle 75* \tan^{-1}(0,5)+12,5 \pi -50 - 12,5\pi - 25= 75* \tan^{-1}(0,5) - 25\approx9,77 }
II način
slika 4
Slika 5
slika 4 i slika 5
75 tan − 1 ( 0 , 5 ) + 12 , 5 π − 12 , 5 π + 25 = 75 tan − 1 ( 0 , 5 ) − 25 ≈ 9 , 77 {\displaystyle 75 \tan^{-1} (0,5) +12,5 \pi - 12,5 \pi +25= 75 \tan^{-1} (0,5) -25 \approx 9,77}
III naćin
slika 6
P = ( ∫ 0 8 100 − x 2 − 5 ) d x − ∫ 0 5 25 − x 2 d x − ∫ 5 8 ( 5 − 25 − ( x − 5 ) 2 d x {\displaystyle P = (\int\limits_{0}^{ 8} \sqrt{100 -x^2} -5)dx - \int\limits_{0}^{ 5} \sqrt{25 -x^2} dx - \int\limits_{5}^{ 8} (5 - \sqrt{25 - (x -5)^2} dx }
∫ 0 5 25 − x 2 d x = ( r = 5 ) = 5 2 π 4 = 25 π 4 {\displaystyle \int\limits_{0}^{ 5} \sqrt{25 -x^2} dx = (r=5) = 5^2 \frac{\pi}{4} = \frac{25 \pi}{4} }
∫ 0 8 ( 100 − x 2 − 5 ) d x = ∫ 0 8 100 − x 2 d x − 40 = ∫ 0 sin − 1 ( 0 , 8 ) 100 − 100 sin 2 θ ( 10 cos θ ) d x − 40 = {\displaystyle \int\limits_{0}^{ 8}( \sqrt{100 -x^2} -5)dx= \int\limits_{0}^{ 8} \sqrt{100 -x^2}dx - 40 = \int\limits_{0}^{ \sin^{-1}(0,8)} \sqrt{100 -100\sin^2 \theta}(10 \cos \theta)dx - 40 = }
100 ∫ 0 sin − 1 ( 0 , 8 ) cos 2 θ d x − 40 = 100 ( θ 2 + sin θ cos θ 2 ) | 0 sin − 1 ( 0 , 8 ) − 40 {\displaystyle 100 \int\limits_{0}^{ \sin^{-1}(0,8)} \cos^2 \theta dx - 40= 100 \bigl( \frac{\theta}{2}+ \frac{ \sin \theta \cos\theta }{2}\bigr)| _0 ^{\sin^{-1}(0,8)} -40 } = 50 sin − 1 ( 0 , 8 ) + 24 − 40 = 50 sin − 1 ( 0 , 8 ) − 16 {\displaystyle 50 \sin^{-1}(0,8) + 24 - 40= 50 \sin^{-1}(0,8) - 16 }
∫ 0 8 5 d x = 40 {\displaystyle \int\limits_{0}^{8} 5dx= 40 }
x = s i n θ {\displaystyle x= sin \theta }
d x = 10 cos θ d θ {\displaystyle dx=10 \cos\theta d\theta } ( 0 ≤ θ ≤ sin − 1 ( 0 , 8 ) ) {\displaystyle (0\leq \theta \leq \sin^ {-1} (0,8)) }
∫ 5 8 ( 5 − 25 − ( x − 5 ) 2 d x = ∫ 5 8 5 d x − ∫ 5 8 ( 25 − ( x − 5 ) 2 ) d x = 15 = − 6 − 12 , 5 sin − 1 ( 0 , 6 ) = 9 − 12 , 5 sin − 1 ( 0 , 6 ) {\displaystyle \int\limits_{5}^{ 8} (5 - \sqrt{25 - (x -5)^2} dx = \int\limits_{5}^{ 8} 5 dx - \int\limits_{5}^{ 8} ( \sqrt{25 - (x -5)^2}) dx = 15 = -6-12,5\sin^{-1}(0,6) = 9- 12,5\sin^{-1}(0,6) }
∫ 5 8 5 d x = 15 {\displaystyle \int\limits_{5}^{ 8} 5 dx = 15 }
∫ 5 8 ( 25 − ( x − 5 ) 2 ) d x = [ x − 5 = u => d x = d u ] = ∫ 0 3 ( 25 − u 2 ) d u = {\displaystyle \int\limits_{5}^{ 8} ( \sqrt{25 - (x -5)^2}) dx =[x-5= u=> dx=du] =\int\limits_{0}^{3} ( \sqrt{25 - u^2}) du = } [ u = 5 sin θ => d u = 5 sin θ d θ {\displaystyle u=5 \sin\theta => du =5\sin\theta d\theta } ]
= − 6 − 12 , 5 sin − 1 ( 0 , 6 ) {\displaystyle = -6-12,5\sin^{-1}(0,6) }
slika 7
P = ( ∫ 0 8 100 − x 2 − 5 ) d x − ∫ 0 5 25 − x 2 d x − ∫ 5 8 ( 5 − 25 − ( x − 5 ) 2 d x = {\displaystyle P = (\int\limits_{0}^{ 8} \sqrt{100 -x^2} -5)dx - \int\limits_{0}^{ 5} \sqrt{25 -x^2} dx - \int\limits_{5}^{ 8} (5 - \sqrt{25 - (x -5)^2} dx = }
= 50 sin − 1 ( 0 , 8 ) − 16 − 25 π 4 − ( 9 − 12 , 5 sin − 1 ( 0 , 6 ) ) = 50 sin − 1 ( 0 , 8 ) − 25 π 4 − 25 + 12 , 5 sin − 1 ( 0 , 6 ) ) = {\displaystyle = 50 \sin^{-1}(0,8) - 16 -\frac{25\pi}{4} - (9- 12,5\sin^{-1}(0,6))= 50 \sin^{-1}(0,8) -\frac{25\pi}{4} -25 + 12,5\sin^{-1}(0,6))= }
[ θ = sin − 1 ( 0 , 6 ) ) ∧ α = sin − 1 ( 0 , 8 ) ) ∧ α + θ = π 2 ) = 12 , 5 sin − 1 ( 0 , 8 ) + 12 , 5 sin − 1 ( 0 , 6 ) = 25 π 4 ] = {\displaystyle [ \theta = \sin^{-1}(0,6)) \land \alpha = \sin^{-1}(0,8))\land \alpha + \theta = \frac{\pi}{2} \Bigr)= 12,5 \sin^{-1}(0,8) + 12,5 \sin^{-1}(0,6) =\frac{25 \pi}{4}]= }
37 , 5 sin − 1 ( 0 , 8 ) + 12 , 5 sin − 1 ( 0 , 8 ) + 12 , 5 sin − 1 ( 0 , 6 ) − 25 − 25 π 4 ] = 37 , 5 sin − 1 ( 0 , 8 ) + 25 π 4 − 25 − 25 π 4 ] = {\displaystyle 37,5 \sin^{-1}(0,8) +12,5 \sin^{-1}(0,8) + 12,5 \sin^{-1}(0,6) -25 -\frac{25 \pi}{4}]= 37,5 \sin^{-1}(0,8) +\frac{25 \pi}{4} -25 -\frac{25 \pi}{4}]= }
37 , 5 sin − 1 ( 0 , 8 ) − 25 ≈ 9 , 77 = 75 tan − 1 ( 0 , 5 ) − 25 {\displaystyle 37,5 \sin^{-1}(0,8) -25 \approx 9,77= 75 \tan^{-1}(0,5)-25 }
sin − 1 ( x ) = 2 tan − 1 ( x 1 + 1 − x 2 ) {\displaystyle \sin^{-1}(x)= 2 \tan^{-1}(\frac{x}{1+\sqrt{1 -x^2}}) }
sin − 1 ( 0 , 8 ) = 2 tan − 1 ( 0 , 64 1 + 1 − 0 , 64 ) = 2 tan − 1 0 , 5 ) {\displaystyle \sin^{-1}(0,8)= 2 \tan^{-1}(\frac{0,64}{1+\sqrt{1 -0,64}})=2\tan^{-1}0,5) }
kvadrat stranice 10.
polukružnice i 
četvrtina kružnice Slika 1
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( Slika 2)
(Slika 3)
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![{\displaystyle \int \limits _{5}^{8}({\sqrt {25-(x-5)^{2}}})dx=[x-5=u=>dx=du]=\int \limits _{0}^{3}({\sqrt {25-u^{2}}})du=}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/21a828581834a2a264197596f345f1462184211c)
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![{\displaystyle [\theta =\sin ^{-1}(0,6))\land \alpha =\sin ^{-1}(0,8))\land \alpha +\theta ={\frac {\pi }{2}}{\Bigr )}=12,5\sin ^{-1}(0,8)+12,5\sin ^{-1}(0,6)={\frac {25\pi }{4}}]=}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/cb82758cebc1945be3c40d687fefaa10d2fc64b9)
![{\displaystyle 37,5\sin ^{-1}(0,8)+12,5\sin ^{-1}(0,8)+12,5\sin ^{-1}(0,6)-25-{\frac {25\pi }{4}}]=37,5\sin ^{-1}(0,8)+{\frac {25\pi }{4}}-25-{\frac {25\pi }{4}}]=}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/3d72a61b72ccc0df065875fff16a9f2cb4ec725b)
