Melody, we can also do this in this manner .
Surface area = 2pi ∫f(x) √[1 + [ f ' (x)]^2 ] dx from a to b
If we rotate the line f(x) = (10/8)x = 1.25x around the x axis from x = 0 to x = 8 we have the same cone laid on its side..... and we have.... {Remember that f 'x = 1.25.....and we must square this under the root }
2 pi ∫ (1.25x ) √[1 + [1.25]^2] dx from 0 to 8 =
2pi √[1 + [1.25]^2] [ 1.25/2] x^2 from 0 to 8 =
1.25pi √[1 + [1.25]^2] * 64 = 402.32 sq units
And we can add the area of the base to this.
