# `y = 3x , 0<=x<=3` Set up and evaluate the definite integral for the area of the surface generated by revolving the curve about the x-axis.

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The quantity to be calculated is the area of what is called a surface of revolution. The function `y = 3x ` is rotated about the x-axis and the surface that is created in this way is a surface of revolution. The area to be calculated is definite, since we consider only the region of the x-axis `x in [0,3] `, that is, `x ` between 0 and 3.

The formula for a surface of revolution (which is an area, A) is given by

`A = int_a^b (2pi y) sqrt(1 + (frac(dy)(dx))^2) dx `

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The circumference of the surface at each point along the x-axis is `2pi y ` and this is added up (integrated) along the x-axis by cutting the function into tiny lengths of `sqrt(1 + (frac(dy)(dx))^2) dx`

ie, the arc length of the function in a segment of the x-axis `dx ` in length, which is the hypotenuse of a tiny triangle with width `dx ` and height `dy `.  These lengths are then multiplied by the circumference of the surface at that point `2 pi y ` to give the surface area of rings around the x-axis that have tiny width `dx ` yet have edges that slope towards or away from the x-axis. The tiny sloped rings are added up to give the full sloped surface area of revolution.

In this case, ` ` `frac(dy)(dx) = 3 ` and since the range over which to take the arc length is `[0,3] ` we have `a = 0 ` and `b = 3 `. Therefore, the area required, A, is given by

`A = int_0^3 6 pi x sqrt(10) dx = 3sqrt(10) pi x^2 | _0^3 = 27sqrt(10)pi `