You may use the following two formulas to evaluate the surface area of an irregular shaped object, such that:

`S = int_a^b 2piy*sqrt(1 + ((dy)/(dx))^2) dx`

`S = int_a^b 2pix*sqrt(1 + ((dx)/(dy))^2) dy`

Considering the following example in which you need to determine the surface area of the following solid of revolution, obtained by rotating `y = sqrt(4 - x^2), x in [-1,1]` , yields:

`S = int_(-1)^1 2piy*sqrt(1 + ((dy)/(dx))^2) dx`

`S = int_(-1)^1 2pi sqrt(4 - x^2)*2/sqrt(4 - x^2) dx`

`S = int_(-1)^1 4pi dx`

`S = 4pi*x|_(-1)^1`

Using the fundamental theorem of calculus yields:

`S = 4pi*(1 - (-1)) => S = 8pi`

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