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

Jan 13, 2016

Determine the sign, then integrate by parts. Area is:

$A = 39.6345$

#### Explanation:

You have to know whether $f \left(x\right)$ is negative or positive in $\left[1 , 3\right]$. Therefore:

$x {e}^{-} x - x {e}^{x}$

$x \left({e}^{-} x - {e}^{x}\right)$

To determine a sign, the second factor will be positive when:

${e}^{-} x - {e}^{x} > 0$

$\frac{1}{e} ^ x - {e}^{x} > 0$

${e}^{x} \cdot \frac{1}{e} ^ x - {e}^{x} \cdot {e}^{x} > {e}^{x} \cdot 0$

Since ${e}^{x} > 0$ for any $x \in \left(- \infty , + \infty\right)$ the unequality doesn't change:

$1 - {e}^{x + x} > 0$

$1 - {e}^{2 x} > 0$

${e}^{2 x} < 1$

$\ln {e}^{2 x} < \ln 1$

$2 x < 0$

$x < 0$

So the function is only positive when x is negative and vice versa. Since there is also an $x$ factor in $f \left(x\right)$

$f \left(x\right) = x \left({e}^{-} x - {e}^{x}\right)$

When one factor is positive, the other is negative, so f(x) is always negative. Therefore, the Area:

$A = - {\int}_{1}^{3} f \left(x\right) \mathrm{dx}$

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

$A = - {\int}_{1}^{3} x {e}^{-} x \mathrm{dx} + {\int}_{1}^{3} x {e}^{x} \mathrm{dx}$

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

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

$A = {\left[x {e}^{-} x\right]}_{1}^{3} - {\int}_{1}^{3} \left(x\right) ' {e}^{-} x \mathrm{dx} + {\left[x \left({e}^{x}\right)\right]}_{1}^{3} - {\int}_{1}^{3} \left(x\right) ' {e}^{x} \mathrm{dx}$

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

$A = {\left[x {e}^{-} x\right]}_{1}^{3} - {\left[- {e}^{-} x\right]}_{1}^{3} + {\left[x \left({e}^{x}\right)\right]}_{1}^{3} - {\left[{e}^{x}\right]}_{1}^{3}$

$A = \left(3 {e}^{-} 3 - 1 \cdot {e}^{-} 1\right) + \left({e}^{-} 3 - {e}^{-} 1\right) + \left(3 {e}^{3} - 1 \cdot {e}^{1}\right) - \left({e}^{3} - {e}^{1}\right)$

$A = \frac{3}{e} ^ 3 - \frac{1}{e} + \frac{1}{e} ^ 3 - \frac{1}{e} + 3 {e}^{3} - e - {e}^{3} + e$

$A = \frac{4}{e} ^ 3 - \frac{2}{e} + 2 {e}^{3}$

Using calculator:

$A = 39.6345$

Area = 11,336.8 square units

#### Explanation:

the given $f \left(x\right) = x {e}^{-} x - x {e}^{x}$

for simplicity let $f \left(x\right) = y$

and $y = x {e}^{-} x - x {e}^{x}$

the first derivative $y '$ is needed in the computation of the surface area.

Area $= 2 \pi {\int}_{1}^{3} y$ $\mathrm{ds}$

where $\mathrm{ds}$$= \sqrt{1 + {\left(y '\right)}^{2}}$ $\mathrm{dx}$

Area $= 2 \pi {\int}_{1}^{3} y$ $\sqrt{1 + {\left(y '\right)}^{2}}$ $\mathrm{dx}$

Determine the first derivative $y '$:

differentiate $y = x \left({e}^{-} x - {e}^{x}\right)$ using the derivative of product formula

$y ' = 1 \cdot \left({e}^{-} x - {e}^{x}\right) + x \cdot \left({e}^{-} x \cdot \left(- 1\right) - {e}^{x}\right)$

$y ' = {e}^{-} x - {e}^{x} - x \cdot {e}^{-} x - x \cdot {e}^{x}$

after simplification and factoring, the result is

the first derivative $y ' = {e}^{-} x \cdot \left(1 - x\right) - {e}^{x} \cdot \left(1 + x\right)$

Compute now the Area:

Area = $2 \pi {\int}_{1}^{3} y$ $\mathrm{ds}$

Area $= 2 \pi {\int}_{1}^{3} y$ $\sqrt{1 + {\left(y '\right)}^{2}}$ $\mathrm{dx}$

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

For complicated integrals like this, we may use Simpson's Rule:

so that

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

Area = -11,336.804

this involves the direction of revolution so that there can be negative surface area or positive surface area. Let us just consider the positive value Area = 11336.804 square units