Because b > a > 0, a given curve rotates around the y axis, which is a ring horizontally placed with the origin as the center. Its volume V is equal to the left half of the circle. The plane figure enclosed by the right half is X = B+√ (A 2-Y 2), Y =-A, Y = A, and the Y axis rotates around the Y axis once, minus the volume of the left half.

V=V 1-V2

=π∫(-a,a)[b+√(a^2-y^2)]^2dy

-π∫(-a,a)[b-√(a^2-y^2)]^2dy

=π∫(-a,a){[b+√(a^2-y^2)]^2-[b-√(a^2-y^2)]^2}dy

=4πb∫(-a,a)√(a^2-y^2)dy

=8πb∫(0,a)√(a^2-y^2)dy.

Let y = asint, then dy = acostdt. When y = 0, t = 0；; When y = a, t = π/2. therefore

V=8πb∫(0，π/2)acost * acostdt

= 8 π a 2b ∫ (0, π/2) cos 2 Tate

=8πa^2b * π/4=2π^2a^2b.