Region R is bounded by x=2-2y^2 and x=1-y^2

Question

find the volume of the solid generated if R is revolved about the y-axis

Anybody help me with this? thanks!

Answer 1

The graphs intersect at (0,-1) and (0,1).
If this area is spun around the y-axis, you can think of the solid as a bunch of thin discs, of thickness dy, each with a hole in the middle.

So, the area is

a = ∫[-1,1] π(R^2-r^2) dy where
R = 2-2y^2 and r = 1-y^2
a = ∫[-1,1] π((2-2y^2)^2-(1-y^2)^2) dy
= 3π∫[-1,1] (y^4-2y^2+1) dy
or, using the symmetry of the figure,
a = 6π∫[0,1] (y^4-2y^2+1) dy
= 6π(1/5 - 2/3 + 1)
= 16π/5

You can also think of the shape as a set of nested shells, each of thickness dx.
x = 2-2y^2, so y = √(1 - x/2)
x = 1-y^2, so y = √(1-x)
We can again exploit the symmetry, but we still have to divide the region into two parts:
for 0<=x<=1, the height of each shell is just the difference in y-values.
For 1<=x<=2, the height is just the y-value, since the x-axis is the lower boundary.

a/2 = ∫[0,1] 2πx(√(1-x/2)-√(1-x)) dx + ∫[1,2] 2πx(√(1-x/2)) dx
= 2π/15 (12-7√2) + 14π/15 √2
= 8π/5
So, a = 16π/5

Answer 2

Great! Except that you wrote "area" instead volume....:P