a)
y=x3
dxdy=3x2
Surface area of revolution
2π∫01x31+(3x2)2 dx
let u=1+9x4
du=36x3 dx
2π∫u36du
18π(32)(u)23
27π(1+9x4)23 (0, 1)
27π((10)23−1)
1.13412π
b)
y=x2
x= y
dydx=2y1
Surface are of revolution
2π∫02y1+(2y1)2 dy
2π∫02y+41 dy
2π (32)(y+41)23 (0, 2)
34π ( (2+41)23−(41)23)
2.6958\pi\
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