Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 153918 by joki last updated on 12/Sep/21

the base of an object is in the form of a circle with  radius 1. suppose that all section of the object are  perpendicular to a diameter of a square. determine  the volume of the object?

$$\mathrm{the}\:\mathrm{base}\:\mathrm{of}\:\mathrm{an}\:\mathrm{object}\:\mathrm{is}\:\mathrm{in}\:\mathrm{the}\:\mathrm{form}\:\mathrm{of}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{with} \\ $$$$\mathrm{radius}\:\mathrm{1}.\:\mathrm{suppose}\:\mathrm{that}\:\mathrm{all}\:\mathrm{section}\:\mathrm{of}\:\mathrm{the}\:\mathrm{object}\:\mathrm{are} \\ $$$$\mathrm{perpendicular}\:\mathrm{to}\:\mathrm{a}\:\mathrm{diameter}\:\mathrm{of}\:\mathrm{a}\:\mathrm{square}.\:\mathrm{determine} \\ $$$$\mathrm{the}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{the}\:\mathrm{object}? \\ $$

Answered by talminator2856791 last updated on 12/Sep/21

       ≡ 2∫_0 ^( (π/2))  (2∙sin(x))^2 dx            = 2π

$$\: \\ $$$$\:\:\:\:\equiv\:\mathrm{2}\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \:\left(\mathrm{2}\centerdot\mathrm{sin}\left({x}\right)\right)^{\mathrm{2}} {dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}\pi \\ $$$$\: \\ $$

Commented by alisiao last updated on 12/Sep/21

=  (2 ∫_0 ^( (𝛑/2)) ( ((e^(ix) −e^(−ix) )/i))^2 dx)    =(−2 ∫_0 ^( (𝛑/2)) ( e^(2ix)  −2 +e^(−2ix) )dx)    =  (−2 ((e^(2ix) /(2i)) − 2x − (e^(−2ix) /(2i)))_( 0) ^( (𝛑/2))  )    =(−2( sin(2x) −2 x)_( 0) ^( (𝛑/2)) )    =  [ −2(0−𝛑)] = 2𝛑    ⟨ M . T  ⟩

$$=\:\:\left(\mathrm{2}\:\int_{\mathrm{0}} ^{\:\frac{\boldsymbol{\pi}}{\mathrm{2}}} \left(\:\frac{\boldsymbol{{e}}^{\boldsymbol{{ix}}} −\boldsymbol{{e}}^{−\boldsymbol{{ix}}} }{\boldsymbol{{i}}}\right)^{\mathrm{2}} \boldsymbol{{dx}}\right) \\ $$$$ \\ $$$$=\left(−\mathrm{2}\:\int_{\mathrm{0}} ^{\:\frac{\boldsymbol{\pi}}{\mathrm{2}}} \left(\:\boldsymbol{{e}}^{\mathrm{2}\boldsymbol{{ix}}} \:−\mathrm{2}\:+\boldsymbol{{e}}^{−\mathrm{2}\boldsymbol{{ix}}} \right)\boldsymbol{{dx}}\right) \\ $$$$ \\ $$$$=\:\:\left(−\mathrm{2}\:\left(\frac{\boldsymbol{{e}}^{\mathrm{2}\boldsymbol{{ix}}} }{\mathrm{2}\boldsymbol{{i}}}\:−\:\mathrm{2}\boldsymbol{{x}}\:−\:\frac{\boldsymbol{{e}}^{−\mathrm{2}\boldsymbol{{ix}}} }{\mathrm{2}\boldsymbol{{i}}}\underset{\:\mathrm{0}} {\overset{\:\frac{\boldsymbol{\pi}}{\mathrm{2}}} {\right)}}\:\right) \\ $$$$ \\ $$$$=\left(−\mathrm{2}\left(\:\boldsymbol{{sin}}\left(\mathrm{2}\boldsymbol{{x}}\right)\:−\mathrm{2}\:\boldsymbol{{x}}\underset{\:\mathrm{0}} {\overset{\:\frac{\boldsymbol{\pi}}{\mathrm{2}}} {\right)}}\right) \\ $$$$ \\ $$$$=\:\:\left[\:−\mathrm{2}\left(\mathrm{0}−\boldsymbol{\pi}\right)\right]\:=\:\mathrm{2}\boldsymbol{\pi} \\ $$$$ \\ $$$$\langle\:\boldsymbol{{M}}\:.\:\boldsymbol{{T}}\:\:\rangle \\ $$

Commented by alisiao last updated on 12/Sep/21

= 8 ∫_0 ^( (𝛑/2))  sin^2 (x) dx = 8 ∫_0 ^( (𝛑/2)) (1/2)(1−cos(2x))dx    = 4 ( x − (1/2)sin(2x))_0 ^( (𝛑/2)) = 4 ( (𝛑/2) ) = 2𝛑    ⟨ M . T  ⟩

$$=\:\mathrm{8}\:\int_{\mathrm{0}} ^{\:\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\boldsymbol{{sin}}^{\mathrm{2}} \left(\boldsymbol{{x}}\right)\:\boldsymbol{{dx}}\:=\:\mathrm{8}\:\int_{\mathrm{0}} ^{\:\frac{\boldsymbol{\pi}}{\mathrm{2}}} \frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\boldsymbol{{cos}}\left(\mathrm{2}\boldsymbol{{x}}\right)\right)\boldsymbol{{dx}} \\ $$$$ \\ $$$$=\:\mathrm{4}\:\left(\:\boldsymbol{{x}}\:−\:\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{sin}}\left(\mathrm{2}\boldsymbol{{x}}\right)\right)_{\mathrm{0}} ^{\:\frac{\boldsymbol{\pi}}{\mathrm{2}}} =\:\mathrm{4}\:\left(\:\frac{\boldsymbol{\pi}}{\mathrm{2}}\:\right)\:=\:\mathrm{2}\boldsymbol{\pi} \\ $$$$ \\ $$$$\langle\:\boldsymbol{{M}}\:.\:\boldsymbol{{T}}\:\:\rangle \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com