# `y = sqrt(9-x^2) , -2<=x<=2` Set up and evaluate the definite integral for the area of the surface generated by revolving the curve about the x-axis.

*print*Print*list*Cite

The function `y = sqrt(9-x^2) ` is rotated about the x-axis and the surface area 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] `, ie `x ` between 0 and 3.

The formula for a surface of revolution A is given by

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

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 infinitessimal 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 `, height `dy `. These lengths are then multiplied by the circumference of the surface at that point, `2pi 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) = -frac(x)(sqrt(9 -x^2)) `

and since the range over which to take the arc length is `[-2,2] ` we have `a = -2 ` and `b=2 ` . Therefore, the area required, A, is given by

`A= int_(-2)^2 2pi sqrt(9-x^2) sqrt(1 + frac(x^2)(9-x^2)) dx `

which can be simplified to

`A = int_(-2)^2 2pi sqrt(9-x^2 + x^2) dx = int_(-2)^2 6 pi dx `

**so that **

`A = 6pi x|_(-2)^2 = 6 pi (2 + 2) = 24pi `