Question

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

Answer 1

I got `22.943`, and so does Wolfram Alpha here.

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The surface area for a revolution around the `x` axis is given by:

`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` is given by:

`((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 `u = -e^(2-x)`. Thus, `du = e^(2-x)dx` and:

`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 `sqrt(a^2 + x^2)`, so let `u = tantheta` to get `du = sec^2thetad theta`. Thus:

`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, `sqrt(1 + e^(4-2x)) - e^(2-x) >=0`, so we can remove the absolute values.

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

Now, we evaluate from `1` to `2`:

`=> [-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)`