Question

Find the area of the surface generated when the given curve is revolved about the given axis. 5x^1/3

Answer 1

`\mathbf{ (\pi)/(675)\Big[ 34√(34) -125\Big] }`

Explanation:

The curve x = f(y)

The area of the surface around the y-axis from y = a → y = b is:

`=\int^b_a 2 \pi x \sqrt{1 + ((dx)/(dy))^2} \ dy`

From the given curve:

`y = (5x)^{^{(1)/(3)}` ; assuming the region bounded by the curve is 0 ≤ y ≤ 1

So;

`y = (5x)^{^{(1)/(3)}`

5x = y³

`x = (1)/(5)y^3`

The differential of the above equation Is:

`(dx)/(dy)= (1)/(5) * (3y^2)`

`(dx)/(dy)= (3)/(5)y^2`

Now, we have the area of the surface produced around the curve
`x = (1)/(5)y^3` through the y axis from the region y = 0 to y = 1

∴

`= \int ^1_0 2 \pi (1)/(5)y^3 \sqrt{1 + ((3)/(5)y^2)^2} \ dy`

`= ( 2 \pi)/(5) \int ^1_0 y^3 \sqrt{1 + (9)/(25)y^4} \ dy`

`= ( 2 \pi)/(5) \int ^1_0 y^3 \sqrt{ (25+9y^4)/(25)} \ dy`

`= ( 2 \pi)/(5) \int ^1_0 y^3 (√(25+9y^4))/(5)} \ dy`

`= ( 2 \pi)/(25) \int ^1_0 y^3 √(25+9y^4)} \ dy`

Let make
`u = √(25+9y^4)`

It implies that:

`u^3 = (25+9y^4)√(25+9y^4)`

`u = √(25+9y^4) \\\\ du = (1)/(2√(25 +9y^4))(36y^3) \ dy`

`du = (18y^3)/(√(25 +9y^4))\ dy`

`y^3dy = (1)/(18)√(25+9y^4) \ du`

when y = 0 ;

`u = √(25+ 9(0)^4)`

`u = √(25)`

u = 5

when y = 1;

`u = √(25+ 9(1)^4)`

`u = √(25+9)`

`u = √(34)`

∴

The equation
`( 2 \pi)/(25) \int ^1_0 y^3 √(25+9y^4)} \ dy` can be written as:

`= (2 \pi)/(25) \int ^(√(34))_(5) (u ) (1)/(18) \ udu`

`= (2 \pi)/(25* 18) \int ^(√(34))_(5) (u ) \ udu`

`= (\pi)/(225) \int ^(√(34))_(5) (u^2 ) \ udu`

`= (\pi)/(225)\Big[ (u^3)/(3) \Big] ^(√(34))_(3)\\`

`\mathbf{= (\pi)/(675)\Big[ 34√(34) -125\Big] }`