Find the volume under z= x^5 + y^5 and above the region bounded by y= x^2 and x= y^2.

Note that volume is not bounded below by y = x² and x = y² (these bound the volume along sides. So volume described is not bounded below at all.

So I guess I'll have to assume that volume is bounded below by xy-plane (z = 0)

The graphs of y = x² and x = y² intersect at point (0,0) and (1,1)
Limits for x: x = 0 to 1
Limits for y: y = x² to √x
Limits for z: z = 0 to (x⁵+y⁵)

V = ∫₀¹ ∫ [x² to √x] ∫ [0 to (x⁵+y⁵)] dz dy dx
V = ∫₀¹ ∫ [x² to √x] (x⁵+y⁵) dy dx
V = ∫₀¹ { (x⁵y + y⁶/6) | [x² to √x] } dx
V = ∫₀¹ { (x⁵√x + x³/6) − (x⁷ + x¹²/6) } dx
V = ∫₀¹ (x^(11/2) + x³/6 − x⁷ − x¹²/6) dx
V = (2/13 x^(13/2) + x⁴/24 − x⁸/8 − x¹³/78) |₀¹
V = 2/13 + 1/24 − 1/8 − 1/78
V = 3/52

Assuming that the lower bound is z = 0:
-----------------
Points of intersection between y = x^2 and x = y^2:
x = (x^2)^2 ==> x(x^3 - 1) = 0 ==> x = 0, 1.

Note that x^(1/2) > x^2 on (0, 1).

Hence, the volume equals
∫(x = 0 to 1) ∫(y = x^2 to x^(1/2)) (x^5 + y^5) dy dx
= ∫(x = 0 to 1) (x^5 y + y^6/6) {for y = x^2 to x^(1/2)} dx
= ∫(x = 0 to 1) [x^(11/2) + (1/6) x^3 - x^7 - (1/6) x^12] dx
= [(2/13) x^(13/2) + (1/24) x^4 - (1/8) x^8 - (1/78) x^13] {for x = 0 to 1}
= 3/52.

I hope this helps!

Intersection points of the curves on the xy-plane:

x^2 = x^4 => x^4 - x^2 = 0 => x = 0, -1(doesn't exist), +1

Volume = ∫[0, 1]∫[x^2, √x] (x^5 + y^5) dy dx

= ∫[0,1] x^5y + (y^6)/6 eval. from x^2 to √x dx

= ∫[0,1] x^(11/2) + (x^3)/6 - x^7 - (x^12)/6 dx

= 2/13*x^(13/2) + (x^4)/24 - (x^8)/8 - (x^13)/78 eval. from 0 to 1

= 2/13 + 1/24 - 1/8 - 1/78 = 2/13 - 1/12 - 1/78 = 11/78 - 1/12 = 3/52