How do you find the area of the surface generated by rotating the curve about the y-axis $x=t+1, y=1/2t^2+t, 0<=t<=2$ ?

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Answer 1 · 2018-06-25T21:09:23

$= (2 pi)/3( 10 sqrt(10) - 2 sqrt(2)) " sq units"$

$bbr(t) = << t+1, 1/2 t(t+2) >>, qquad 0<=t<=2$

$ds = sqrt(dot x^2 + dot y^2) \ dt$

$dS = ds * 2 pi x = 2 pi (t+1) sqrt((1)^2 + (t+1)^2) \ dt$

$= 2 pi (t+1) sqrt(t^2 + 2t + 2) \ dt$

$S =2 pi int_0^2 dt qquad (t+1) sqrt(t^2 + 2t + 2)$

$=2 pi int_0^2 dt qquad d/dt ( 1/3(t^2 + 2t + 2)^(3/2))$

$=(2 pi)/3 [ (t^2 + 2t + 2)^(3/2)]_0^2$

$=(2 pi)/3 ( 10^(3/2) - 2^(3/2))$

$= (2 pi)/3( 10 sqrt(10) - 2 sqrt(2)) approx 60.3 " sq units"$

How do you find the area of the surface generated by rotating the curve about the y-axis x=t+1, y=1/2t^2+t, 0<=t<=2x=t+1, y=1/2t^2+t, 0<=t<=2?