Find the volume of the solid generated by revolving the region about the given line.

Question

The region in the second quadrant bounded above by the curve y = 16 - x2, below by the x-axis, and on the right by the y-axis, about the line x = 1

I have gathered, that washer method is to be used - (-4,0) is the shaded area.

Answer 1

washers is a good way:
v = ∫[0,16] π(R^2-r^2) dy
where R=1+|x|, r=1
v = π∫[0,16] (1+√(16-y))^2 - 1 dy
= -π(4/3 (16-y)^(3/2) + 1/2 (16-y)^2) [0,16]
= 640π/3

Or, using shells,
v = ∫[-4,0] 2πrh dx
where r=1-x and h=y
v = 2π∫[-4,0] (x+1)(16-x^2) dx
= 2πx (x^3/4 - x^2/3 - 8x + 16)[-4,0]
= 640π/3

Hmm! shells is less complicated this time.

Answer 2

Where does the 4/3 and 3/2 come from?

Thank you

Answer 3

∫(1+√(16-y))^2 - 1 dy
let u = 16-y
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