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

##### 1 Answer

I got

The **surface area** for a revolution around the

#S = 2 pi int_(a)^(b) f(x)sqrt(1 + ((dy)/(dx))^2)dx# (which is basically a projection of the circumference along the function

#f(x)# whose arc length you could have found.)

In this case,

#((dy)/(dx))^2 = (-e^(2-x))^2#

And so we have:

#S = 2 pi int_1^2 e^(2-x) sqrt(1 + (-e^(2-x))^2)dx#

First, let

#S = 2 pi int sqrt(1 + u^2)du#

where we omit the integral bounds for now. Then we can see it looks like the form

#S = 2 pi int sqrt(1 + tan^2theta)sec^2thetad theta#

#= 2 pi int sec^3thetad theta#

And you should have written down this integral in class to be:

#= 2 pi overbrace([1/2(secthetatantheta + ln|sectheta + tantheta|)])^(int sec^3 theta d theta)#

#= pi secthetatantheta + pi ln|sectheta + tantheta|# (if you didn't, then you may have unknowingly asked a difficult question!)

Now, back-substitute.

#=> pi usqrt(1 + u^2) + pi ln|sqrt(1 + u^2) + u|#

#= pi usqrt(1 + (-e^(2-x))^2) + pi ln|sqrt(1 + (-e^(2-x))^2) + (-e^(2-x))|#

#= -pi e^(2 - x)sqrt(1 + e^(4-2x)) + pi ln|sqrt(1 + e^(4-2x)) - e^(2-x)|#

As it turns out,

#= -pi e^(2 - x)sqrt(1 + e^(4-2x)) + pi ln(sqrt(1 + e^(4-2x)) - e^(2-x))#

Now, we evaluate from

#=> [-pi e^(2 - 2)sqrt(1 + e^(4-2*2)) + pi ln(sqrt(1 + e^(4-2*2)) - e^(2-2))] - [-pi e^(2 - 1)sqrt(1 + e^(4-2*1)) + pi ln(sqrt(1 + e^(4-2*1)) - e^(2-1))]#

#= [-pi sqrt(2) + pi ln(sqrt(2) - 1)] - [-pi esqrt(1 + e^2) + pi ln(sqrt(1 + e^2) - e)]#

#= -pi sqrt(2) + pi ln(sqrt(2) - 1) + pi esqrt(1 + e^2) - pi ln(sqrt(1 + e^2) - e)#

#~~# #color(blue)(22.943)#