What is the surface area of the solid created by revolving #f(x) = 1/(x+e^x) , x in [3,4]# around the x axis?

Question

What is the surface area of the solid created by revolving #f(x) = 1/(x+e^x)  , x in [3,4]# around the x axis?

Aurora Alvarado · Accepted Answer

show below

the surface area produce from revolving =volume

these is the sketch of function

these is the are area wanted to calculate between x=3 and x=4

the volume produced from revolving the surface area between x=3 and x=4 is calculated by ;

#v=piint_3^4[f(x)]^2*dx#
Then complete the steps normally.

Charles Arocha · Answer

undefined To find the surface area of the solid created by revolving $ f(x) = \frac{1}{x + e^x} $, where $ x $ ranges from 3 to 4, around the x-axis, we can use the formula for the surface area of a solid of revolution:

$$ S = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} \, dx $$

where $ f'(x) $ is the derivative of $ f(x) $.

1. Find $ f'(x) $:
$$ f'(x) = -\frac{1}{(x + e^x)^2} + \frac{d}{dx}(e^x) = -\frac{1}{(x + e^x)^2} + e^x $$

2. Plug $ f(x) $ and $ f'(x) $ into the formula for surface area:
$$ S = \int_{3}^{4} 2\pi \frac{1}{x + e^x} \sqrt{1 + \left(-\frac{1}{(x + e^x)^2} + e^x\right)^2} \, dx $$

3. Compute the integral numerically to find the surface area.