Задача 35913 № 9.3.235 Кривая вращается вокруг оси....

bolanebyla

2019-04-16 14:09:49

№ 9.3.235 Кривая вращается вокруг оси. Вычислить площадь поверхности вращения

sova

2019-04-16 14:44:58

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S_(поверхности вращения вокруг оси ОХ)=2π ∫ ^(t_(2))_(t_(1))y(t)*sqrt((x`(t))^2+(y`(t))^2)dt

x`(t)= - 2Rsint + 2Rsin2t
y`(t)=2Rcost - 2Rcos2t

(x`(t))^2= 4R^2sin^2t -8R^2sint*sin2t+4R^2sin^22t
(y`(t))^2=4R^2cos^t -8R^2cost*cos2t+4R^2cos^22t


(x`(t))^2+(y`(t))^2=4R^2*(sin^2t+cos^2t+sin^22t+cos^22t-2(cost*cos2t+sint*sin2t))

(x`(t))^2+(y`(t))^2=4R^2*(1+1-2*cos(2t-t))=8R^2*(1-cost)=

=16R^2sin^2(t/2)

sqrt(16R^2sin^2(t/2))=4Rsin(t/2)

[b]S_(поверхности вращения вокруг оси Ох)=
= 2π ∫^(2π)_(0)(2Rsint-Rsin2t)*4Rsin(t/2)dt[/b]=

=8πR^2 ∫^(2π)_(0)(2sint*sin(t/2)-sin2t*sin(t/2))dt=

=8πR^2 ∫^(2π)_(0)(sin(3t/2)+sin(t/2)-(1/2)sin(5t/2)- (1/2)sin(3t/2))dt=

=8πR^2 ∫^(2π)_(0)((1/2)sin(3t/2)+sin(t/2)-(1/2)sin(5t/2))dt=

=8πR^2*((-1/2)*(2/3)cos(3t/2)) -2cos(t/2)+(1/2)*(2/5)cos(5t/2))^(2π)_(0)=

=8πR^2*((-1/3)cos3π+(1/3)cos0 -2cosπ+2cos0 +(1/5)cos(5π)-(1/5)cos0)=

=8πR^2*((2/3)+4-(2/5))=8πR^2*(64/15)= [b]512πR^2/15[/b]