Question

Find the area of the surface generated when the given curve is revolved about the y-axis. The part of the curve y=4x-1 between the points (1, 3) and (4, 15)

Answer 1

Explanation:

Let take a look at the given function y = 4x - 1 whose point is located between (1,3) and (4,15) on the graph.

Here, the function of y is non-negative. Now, expressing y in terms of x in y = 4x- 1

4x = y + 1

`x = (y+1)/(4)`

`x = (1)/(4)y + (1)/(4)`

By integration, the required surface area in the revolve is:

`S = \int^(15)_( 3) 2 \pi g (y) √(1+g'(y^2) \ dy )`

where;

g(y) =
`x = (1)/(4)y + (1)/(4)`

∴

`S = \int^(15)_( 3) 2 \pi \Big( (1)/(4)y + (1)/(4)\Big) \sqrt{1+\Bigg(\Big( (1)/(4)y + (1)/(4)\Big)'\Bigg)^2 \ dy }`

`S = (1)/(2) \pi \int^(15)_( 3) (y+1) \sqrt{1+\Bigg(\Big( (1)/(4)\Big ) \Bigg)^2 \ dy } \\ \\ \\ S = (1)/(2) \pi \int^(15)_( 3) (y+1) (√(17))/(4) \ dy`

`S = (√(17))/(8) \pi \int^(15)_( 3) (y+1) \ dy`

`S = (√(17) \pi)/(8) ((1)/(2)(y+1)^2)\Big|^(15)_(3) \\ \\ S = (√(17) \pi)/(8) ((1)/(2)(15+1)^2-(1)/(2)(3+1)^2 ) \\ \\ S = (√(17) \pi)/(8) *120 \\ \\\mathbf{ S = 15 √(17)x}`