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

Mar 29, 2018

The surface area is approximately 105.75.

#### Explanation:

The surface area $S$ is given by

$S = 2 \pi {\int}_{1}^{3} x {e}^{-} x - {e}^{x} \mathrm{dx} = 2 \pi \left[{\int}_{1}^{3} x {e}^{-} x \mathrm{dx} - {\int}_{1}^{3} {e}^{x} \mathrm{dx}\right]$.

$\int x {e}^{-} x = - {e}^{-} x \left(x + 1\right) + C$

which can be obtained by integration by parts, so

$S = 2 \pi \left[- {e}^{-} x \left(x + 1\right) - {e}^{x}\right]$ evaluated from 1 to 3.

$S = 2 \pi \left[- {e}^{-} 3 \cdot 4 - {e}^{3} + {e}^{-} 1 \cdot 2 + e\right] \approx - 105.75$

To make this physically meaningful, we would just take the absolute value of this and say that

$S \approx 105.75$.

The volume, $V$, can be calculated from $\pi {\int}_{1}^{3} {\left(x {e}^{-} x - {e}^{x}\right)}^{2} \mathrm{dx}$.

$\pi {\int}_{1}^{3} {\left(x {e}^{-} x - {e}^{x}\right)}^{2} \mathrm{dx} = \pi {\int}_{1}^{3} {x}^{2} {e}^{- 2 x} - 2 x + {e}^{2 x} \mathrm{dx}$

$= \pi \left({\int}_{1}^{3} {x}^{2} {e}^{- 2 x} \mathrm{dx} - 2 {\int}_{1}^{3} x \mathrm{dx} + {\int}_{1}^{3} {e}^{2 x} \mathrm{dx}\right)$

The tough one here is $\int {x}^{2} {e}^{- 2 x} \mathrm{dx}$. This is done by integration by parts. If you need me to show you the steps for you, just indicate in the comments. For now, I will simply note that

$\int {x}^{2} {e}^{- 2 x} \mathrm{dx} = - \frac{1}{4} {e}^{- 2 x} \left(2 {x}^{2} + 2 x + 1\right) + C$

And so we have

$V = \pi \left[- \frac{1}{4} {e}^{- 2 x} \left(2 {x}^{2} + 2 x + 1\right) - {x}^{2} + {e}^{2 x} / 2\right]$ evaluated from 1 to 3.

$V = \pi \left[- {e}^{- 6} / 4 \cdot 25 - 9 + {e}^{6} / 2 + {e}^{- 2} / 4 \cdot 5 + 1 - {e}^{2} / 2\right] \approx 597.45$