Calculus...?

Question

The region bounded by y = 2 + sin x, y = 0, x = 0, and 2pi is revolved about the y-axis. Find the volume that results.

Hint: ∫x sin x dx = sin x - x cos x + C

Volume of the solid of revolution:_________

Answer 1

I didn't find the hint very useful.

In the region from y = 2 to y = 3 (x = 0 to x = π), for a particular value of y we get two values of x, namely
x = arcsin (y-2), x = π - arcsin (y-2)
of which the second is greater. So revolving this around the y-axis gives us
V = π∫(2 to 3)((π - arcsin (y-2))^2 - (arcsin (y-2))^2) dy
= π∫(2 to 3)(π^2 - 2πarcsin (y-2)) dy
= π^3 - 2π^2∫(0 to 1) arcsin y dy
where I've changed variables from y-2 to y to simplify things (if you find this confusing, make a substitution u = y-2 and you should get the same integral as here with y replaced by u).

Now you can verify that ∫arcsin x dx = x arcsin x + √(1-x^2), so we get
V = π^3 - 2π^2[y arcsin y + √(1-y^2)][0 to 1]
= π^3 - 2π^2 [(1.π/2 + 0) - (0 + 1)]
= 2π^2.

Similarly, between y=1 and y=2 we have two values for x, which are π - arcsin (y-2) and 2π + arcsin (y-2) this time. So we get a similar integral:
V = π∫(1 to 2) ((2π + arcsin (y-2))^2 - (π - arcsin (y-2))^2) dy
= π∫(1 to 2) (3π^2 + 6π arcsin (y-2)) dy
= 3π^3 + 6π^2∫(-1 to 0) arcsin y dy
= 3π^3 + 6π^2 [y arcsin y + √(1-y^2)][-1 to 0]
= 3π^3 + 6π^2 [(0 + 1) - (-1(-π/2) + 0)]
= 6π^2.

So the total volume is 2π^2 + 6π^2 = 8π^2.

Answer 2

Hmmm....Mr. Manuka's solution is incorrect.