What is the surface area of the solid created by revolving f(x) =2x+5 , x in [1,2] around the x axis?

1 Answer
Jan 5, 2017

We need to calculate the area of te curved surface and the two end circles.

Explanation:

The end circles have radii f(1)=7 and f(2)=9 respectively and so have areas 49pi and 81pi.

The are of the curved surface may be calculated by considering an infinitesimal ring of radius f(x) between the values x and x+dx, which will have thickness dl=sqrt(df^2(x)+dx^2) and therefore Area=2pif(x)dl. The total curved area will then be \int_1^2 2pif(x)dl
Now, dl=sqrt(df^2(x)+dx^2)=dxsqrt(((df(x))/dx)^2+1)=dxsqrt(2^2+1)
=dxsqrt5.
So we have to evaluate the integral
\int_1^2 dx 2pi f(x) sqrt5=2sqrt5pi [2x^2/2+5x]_1^2=2sqrt5pi [14-6]
=16sqrt5pi.

The total area is then =(16sqrt5+49+81)pi=(16sqrt5+130)pi.