求 x = e t − t , y = 4 e t 2 , 0 < t < 1 {\displaystyle x=e^{t}-t,\;y=4e^{\frac {t}{2}},\;0<t<1} 绕 y 轴的表面积
A = ∫ C 2 π r d S = ∫ 2 π x d x 2 + y 2 = ∫ 2 π x ( d x d t ) 2 + ( d y d t ) 2 d t = ∫ 0 1 2 π x ( e t − 1 ) 2 + ( 2 e t 2 ) 2 d t = ∫ 0 1 2 π x e 2 t − 2 e t + 1 + 4 e t d t = ∫ 0 1 2 π x e 2 t + 2 e t + 1 d t = ∫ 0 1 2 π ( e t − t ) ( e t + 1 ) d t = 2 π ∫ 0 1 e 2 t + e t − t e t − t d t = 2 π [ e 2 t 2 + e t − ( t e t − e t ) − t 2 2 ] 0 1 = 2 π [ e 2 2 + e − 1 2 − ( 1 2 + 1 + 1 ) ] = π ( e 2 + 2 e − 6 ) {\displaystyle {\begin{aligned}&A=\int _{C}2\pi rdS=\int 2\pi xd{\sqrt {x^{2}+y^{2}}}\\&=\int 2\pi x{\sqrt {\left({\frac {dx}{dt}}\right)^{2}+\left({\frac {dy}{dt}}\right)^{2}}}dt\\&=\int _{0}^{1}2\pi x{\sqrt {\left(e^{t}-1\right)^{2}+\left(2e^{\frac {t}{2}}\right)^{2}}}dt\\&=\int _{0}^{1}2\pi x{\sqrt {e^{2t}-2e^{t}+1+4e^{t}}}dt\\&=\int _{0}^{1}2\pi x{\sqrt {e^{2t}+2e^{t}+1}}dt\\&=\int _{0}^{1}2\pi \left(e^{t}-t\right)\left(e^{t}+1\right)dt\\&=2\pi \int _{0}^{1}e^{2t}+e^{t}-te^{t}-t\,dt\\&=2\pi \left[{\frac {e^{2t}}{2}}+e^{t}-\left(te^{t}-e^{t}\right)-{\frac {t^{2}}{2}}\right]_{0}^{1}\\&=2\pi \left[{\frac {e^{2}}{2}}+e-{\frac {1}{2}}-\left({\frac {1}{2}}+1+1\right)\right]=\pi \left(e^{2}+2e-6\right)\end{aligned}}}