Calculus Answer Wrong

Find the volume bounded by the region y = x^1/3 y = x around y=1 intersections: -1, to 0 : 0 to 1

When I work out this problem, no matter what way I solve it, I get 0 Pie. The book solution is pie (3.14). Could someone work out this problem, and attempt to pinpoint where I'm making the mistake? Again, my answer may actually be correct? Thanks for any replies

From -1 to zero, the line is above the cubic, and from 0 to 1 the cubic is above the line.

You will need to use washer method, pi*R^2 -pi*r^2

R= 1-y, since you are rotating about the line y=1

On (-1,0) R = 1-x^1/3, r= 1-x
On (0,1) R = 1-x , r= 1-x^1/3

INT [ pi( 1-x^(1/3))^2 - pi(1-x)^2 dx on (-1,0) + INT[ pi(1-x)^2 -pi(1-x^(1/3))^2] dx on (0,1)
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On (-1,0):

pi* INT (1-2x^(1/3) + x^(2/3)-1+2x-x^2) dx on (-1,0)

= pi* [ (-3/2)x^(4/3)+ (3/4)x^(4/3) +x^2 -x^3/3]| (-1,0)

= pi[ 0-( -3/2-3/5+1+1/3)]

=23pi/30
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On (0,1) :

Pi* INT[ 1-2x+x^2-1+2x^(1/3)-x^(2/3)] dx (0,1)

= pi*( -x^2+x^3/3 + (3/2)x^(4/3) -(3/5)x^(5/3)] |(0,1)

= pi( -1+1/3+3/2-3/5)-0

=7pi/30

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Total volume = 23pi/30+ 7 pi/30 = pi

Hoping this helps!