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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#?

1 Answer
Jun 25, 2018

Answer:

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

Explanation:

#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"#