calculus
posted by Tiffany .
Find the volume of the solid generated by revolving the region about the given line.
The region in the second quadrant bounded above by the curve y = 16  x2, below by the xaxis, and on the right by the yaxis, about the line x = 1
I have gathered, that washer method is to be used  (4,0) is the shaded area.

washers is a good way:
v = ∫[0,16] π(R^2r^2) dy
where R=1+x, r=1
v = π∫[0,16] (1+√(16y))^2  1 dy
= π(4/3 (16y)^(3/2) + 1/2 (16y)^2) [0,16]
= 640π/3
Or, using shells,
v = ∫[4,0] 2πrh dx
where r=1x and h=y
v = 2π∫[4,0] (x+1)(16x^2) dx
= 2πx (x^3/4  x^2/3  8x + 16)[4,0]
= 640π/3
Hmm! shells is less complicated this time. 
Where does the 4/3 and 3/2 come from?
Thank you 
∫(1+√(16y))^2  1 dy
let u = 16y
du = dy
∫(1+√u)^2  1 du
∫1 + 2√u + u  1
∫2√u + u
2(2/3)u^(3/2) + 1/2 u^2
because
∫u^n = 1/(n+1) u^(n+1) where n=1/2