What is the surface area of the solid created by revolving #f(x)=e^(x+1)/(x+1)# over #x in [0,1]# around the x-axis?

Question

William Abdalla · Accepted Answer

The area of the solid of revolution around the x-axis of a curve is calculated as:

#pi int_a^b f^2(x)dx# #V= pi int_0^1((e^(x+1))/(x+1))^2dx = pi int_0^1e^(2(x+1))/(x+1)^2dx# Substitute #t=2(x+1); dx=dt/2# #V=2pi int_2^4e^t/t^2dt# We can calculate this integral by parts: #int e^t/t^2dt = -int e^t d(t^(-1)) = -e^t/t + int e^t/t # # int e^t/tdt # is not easily solved, you can find it on manuals: # int e^t/tdt =log|x| + sum_0^oox^n/(n*n!)#, so #V=2pi(-e^4/4+e^2/2+log 4 -log 2 +sum_0^oo4^n/(n*n!)-sum_0^oo2^n/(n*n!))#

Ezekiel Alexander · Answer

undefined To find the surface area of the solid created by revolving the function $ f(x) = \frac{e^{x+1}}{x+1} $ over $ x $ in the interval $[0,1]$ around the x-axis, follow these steps:

1. Calculate the arc length of the curve from $ x = 0 $ to $ x = 1 $ using the formula for arc length:

$$ S = \int_{a}^{b} 2\pi y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx $$

2. Determine the derivative $ \frac{dy}{dx} $ of the function $ f(x) $.
3. Evaluate $ y = f(x) $ and $ \frac{dy}{dx} $ at each point in the interval $[0,1]$.
4. Plug the values into the arc length formula and integrate over the interval $[0,1]$.
5. The result will be the surface area of the solid generated by revolving the function around the x-axis.

Let's go through the steps:

1. Calculate the derivative of $ f(x) $:

$$ f(x) = \frac{e^{x+1}}{x+1} $$

$$ f'(x) = \frac{e^{x+1}(x+1) - e^{x+1}}{(x+1)^2} = \frac{e^{x+1}(x) - e^{x+1}}{(x+1)^2} $$

2. Plug the function and its derivative into the arc length formula:

$$ S = \int_{0}^{1} 2\pi \left(\frac{e^{x+1}}{x+1}\right) \sqrt{1 + \left(\frac{e^{x+1}(x) - e^{x+1}}{(x+1)^2}\right)^2} dx $$

3. Evaluate this integral numerically to find the surface area.