I love using triple integrals for volume 
So I'd do theta = 0 to 2pi, r = 3/8 to 5/8, x = 0 to 5/8 - r
ʃʃʃ r dx dr dtheta
ʃʃ (5/8 - r)r dr dtheta
ʃ [5/16r^2 - 1/3r^3]_{r=3/8}^{5/8} dtheta
ʃ 0.0143229167 dtheta
0.0899935396
Obviously your method has rounding issues, it's not actually 0.09 but 11/384 pi, which is a little lower.
edit: heh, stole your integral sign.
So I'd do theta = 0 to 2pi, r = 3/8 to 5/8, x = 0 to 5/8 - r
ʃʃʃ r dx dr dtheta
ʃʃ (5/8 - r)r dr dtheta
ʃ [5/16r^2 - 1/3r^3]_{r=3/8}^{5/8} dtheta
ʃ 0.0143229167 dtheta
0.0899935396
Obviously your method has rounding issues, it's not actually 0.09 but 11/384 pi, which is a little lower.
edit: heh, stole your integral sign.