# Find the volume of the solid formed by rotating the region enclosed by y=e^(5x)+4 , \ y=0, \ x=0, \ x=0.8 about the y-axis. Use integrals to find the volume.

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### 1 Answer

You should use the following formula to find the volume of solid generated rotating the given curve about y axis such that:

`V = pi*int_a^b r^2(y) dy`

The problem provides the equation that defines the curve such that:

`y=e^(5x)+4 => y - 4 = e^(5x) => ln e^(5x) = ln(y - 4) => x = (ln(y - 4))/5`

Hence, evaluating the radius of solid of revolution yields:

`r(y) = (ln(y - 4))/5`

Squaring the radius yields:

`r^2(y) = (ln^2(y - 4))/25`

Hence, you may evaluate the volume such that:

`V = pi*int_5^(e^4+4)(ln^2(y - 4))/25 dy`

`V= (pi/25)int_5^(e^4+4) (ln^2(y - 4)) dy`

You should use the folowing substitution such that:

`y - 4 = u => dy = du`

`V =(pi/25)int_1^(e^4)ln^2 u du`

You should use integration by parts formula such that:

`int fg' = fg - int f'g`

`f = ln^2 u => f' = (2ln u)/u`

`g' = du => g = u`

`int_1^(e^4) ln^2 u du = uln^2 u|_1^(e^4) - int_1^(e^4) u*(2ln u)/u du`

`int_1^(e^4) ln^2 u du = u ln^2 u|_1^(e^4) - 2int_1^(e^4) ln u du`

Using parts again yields:

`f = ln u => f ' = 1/u`

`g' = du => g = u`

`int_1^(e^4) ln^2 u du = u ln^2 u|_1^(e^4) - 2(u*ln u - int_1^(e^4) u*(1/u)du)`

`int_1^(e^4) ln^2 u du = u ln^2 u|_1^(e^4) - 2(u*ln u - u)|_1^(e^4) `

`int_1^(e^4) ln^2 u du = 16e^4 - 8e^4 = 8e^4`

**Hence, evaluating the volume of solid of revolution, under the given conditions, yields `V = (8pi*e^4)/25` .**