FIND THE SURFACE AREA OF THE OBJECT OBTAINED BY ROTATING Y=4+3x2, 1<=x<=2 ABOUT THE Y-AXIS
Surface Integral = 2π∫12x1+[f′(x)]2dx
f(x)=y=4+3x2
f′(x)=6x
Surface Area = 2π∫12x1+(6x)2dx
2π∫12x1+36x2dx
let u = 1+36x2 , dxdu=72x , dx=72xdu
Surface Area =2π∫1272xxudu
=722π∫12udu
=36π.32u3∣12
=54πu3∣12
=54π(1+36x2)3∣12
=[54π(1+36(2)2)3]−[54π(1+36(1)2)3]
=[54π(145)3]−[54π(37)3]
=54π[(145)3−(37)3]
Surface Area =54π[145145−3737]
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