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#.