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Calculate the surface area of the solid of revolution obtained by revolving the region between fx=2x3and the x-axis on 0,3around the x-axis.

Short Answer

Expert verified

The surface area of the solid of revolution obtained by revolving the region between fx=2x3and the x-axis on 0,3around the x-axis is, 1526.81(approximately) square units.

Step by step solution

01

Step 1. Given information

The region betweenfx=2x3and the x-axis on0,3.

02

Step 2. The formula for the surface area is,

Surface area=2abfx1+f'x2dx.

Let fx=2x3,f'x=6x2.

Substitute the known values in the formula.

Surface area=2032x31+6x22dx

role="math" localid="1648890749974" =2032x31+36x4dx

Let u=1+36x4,du=144x3.

role="math" localid="1648892010258" 2x=0x=32x31+36x4dx=22144x=0x=3udu=3623u3203=36231+36x43203=541+36x603=541+3636-1+3606=541+26,244-1=4861526.81

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