What is the surface area of the solid created by

Jamison Rios

Jamison Rios

Answered question

2022-07-03

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

Answer & Explanation

Jordan Mcpherson

Jordan Mcpherson

Beginner2022-07-04Added 16 answers

If you consider the interval (x, x+dx) the length of the arc element of the curve
y=f(x) is:
d l = d x 2 + d y 2 = d x 2 + f ( x ) 2 d x 2 = d x 1 + f ( x ) 2
so the area of the surface element generated by the rotation around the x-axis is:
d S = 2 π y d l = 2 π f ( x ) 1 + f ( x ) 2 d x
Integrating over the interval, the surface is then calculated as:
S = 2 π 1 2 f ( x ) 1 + f ( x ) 2 d x
As f'(x)=1:
S = 2 2 π 1 2 ( x 1 ) d x = 2 π [ ( x 1 ) 2 ] 1 2 = 2 π

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