What is the surface area of the solid created by revolving f ( x ) = 1

Devin Anderson

Devin Anderson

Answered question

2022-06-07

What is the surface area of the solid created by revolving f ( x ) = 1 x 2 for x [ 1 , 2 ] around the x-axis?

Answer & Explanation

Leland Ochoa

Leland Ochoa

Beginner2022-06-08Added 25 answers

The first step is to check to make sure that f(x) is defined on the interval [1,2]. The only discontinuity of the the function is at x=0, so we do not have to concern ourselves with this.
The surface area of a curve on [a,b] rotated around the x-axis is defined by
S = 2 π a b f ( x ) 1 + ( f ( x ) ) 2 d x . The derivative of f(x) is f ( x ) = 2 x 3 . This
means that ( f ( x ) ) 2 = ( 2 x 3 ) 2 = 4 x 6
S = 2 π 1 2 1 x 2 1 + 4 x 6 d x
S = 2 π 1 2 1 x 2 x 6 + 4 x 6 d x
S = 2 π 1 2 1 x 5 x 6 + 4 d x
This integral has no elementary solution.
But according to Wolfram Alpha, the surface area is approximately 4.46 square units.

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