Question

Find the area of the surface generated when the given curve is revolved about the x-axis y=4x+5 [0,2 ]a. 36√17.xb. 36πc. 36π/√17d. 32√17.π

Answer 1

The area of this revolted surface is 36π

Explanation:

To obtain the area of a revolted surface, you have to define:

1) which is the axis on which the surface is revolted: this defines the limits on that axis or hight of the surface. In this case x∈[0;2]

2) which is the expression of the radius of the revolted surface and its dependence with the hight. In this case, the radius expression could be Y=4x+5

3) Define the angular variable: If this is a fully revolted surface, the angular variable will go from 0 to 2π

Now we can obtain the area with a double integral:

`A=\int\limits^(2)_0 { \int\limits^(2\pi)_0 {r} \, d \varphi } \, dx =\int\limits^(2)_0 { \int\limits^(2\pi)_0 {4x+5} \, d \varphi } \, dx =\int\limits^(2)_0 { (2\pi)(4x+5)} \, dx=36\pi`