Question

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?

Answer 1

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

`A=39.6345`

Explanation:

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

`xe^-x-xe^x`

`x(e^-x-e^x)`

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

`e^-x-e^x>0`

`1/e^x-e^x>0`

`e^x*1/e^x-e^x*e^x>e^x*0`

Since `e^x>0` for any `x in(-oo,+oo)` the unequality doesn't change:

`1-e^(x+x)>0`

`1-e^(2x)>0`

`e^(2x)<1`

`lne^(2x) < ln1`

`2x<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(x)`

`f(x)=x(e^-x-e^x)`

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

`A=-int_1^3f(x)dx`

`A=-int_1^3(xe^-x-xe^x)dx`

`A=-int_1^3xe^-xdx+int_1^3xe^xdx`

`A=-int_1^3x*(-(e^-x)')dx+int_1^3x(e^x)'dx`

`A=int_1^3x*(e^-x)'dx+int_1^3x(e^x)'dx`

`A=[xe^-x]_1^3-int_1^3(x)'e^-xdx+[x(e^x)]_1^3-int_1^3(x)'e^xdx`

`A=[xe^-x]_1^3-int_1^3e^-xdx+[x(e^x)]_1^3-int_1^3e^xdx`

`A=[xe^-x]_1^3-[-e^-x]_1^3+[x(e^x)]_1^3-[e^x]_1^3`

`A=(3e^-3-1*e^-1)+(e^-3-e^-1)+(3e^3-1*e^1)-(e^3-e^1)`

`A=3/e^3-1/e+1/e^3-1/e+3e^3-e-e^3+e`

`A=4/e^3 -2/e+2e^3`

Using calculator:

`A=39.6345`

Answer 2

Area = 11,336.8 square units

Explanation:

the given `f(x)=xe^-x -xe^x`

for simplicity let `f(x)=y`

and `y=xe^-x -xe^x`

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

Area `= 2pi int_1^3 y` `ds`

where `ds` `=sqrt(1+ (y')^2)` `dx`

Area `= 2pi int_1^3 y` `sqrt(1+ (y')^2)` `dx`

Determine the first derivative `y'`:

differentiate `y=x(e^-x - e^x)` using the derivative of product formula

`y' = 1*(e^-x-e^x)+x*(e^-x *(-1)-e^x)`

`y'=e^-x - e^x -x*e^-x -x*e^x`

after simplification and factoring, the result is

the first derivative `y'=e^-x*(1-x)-e^x*(1+x)`

Compute now the Area:

Area = `2 pi int_1^3 y` `ds`

Area `= 2pi int_1^3 y` `sqrt(1+ (y')^2)` `dx`

Area
`= 2pi int_1^3 x(e^-x - e^x)` `sqrt(1+ (e^-x*(1-x)-e^x*(1+x))^2` `dx`

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

so that

Area
`= 2pi int_1^3 x(e^-x - e^x)` `sqrt(1+ (e^-x*(1-x)-e^x*(1+x))^2` `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