I need help getting the volume of a solid through integration?

Question

.Determine the volume of the solid lying
under the graph of
z = 5x + 4y
and above the bounded region enclosed by the
graphs of y^2 = x and y = x^2.

Answer 1

The graphs intersect at (0, 0) and (1, 1); so we have
∫(0 to 1) ∫(x^2 to √x) ∫(0 to 5x + 4y) dz dy dx
= ∫(0 to 1) ∫(x^2 to √x) (5x + 4y) dy dx
= ∫(0 to 1) [5xy + 2y^2] [x^2 to √x] dx
= ∫(0 to 1) (5x(√x - x^2) + 2(x - x^4)) dx
= ∫(0 to 1) (5x^(3/2) - 5x^3 + 2x - 2x^4) dx
= [5x^(5/2).(2/5) - 5x^4 / 4 + x^2 - 2x^5 / 5] [0 to 1]
= (5(2/5) - 5/4 + 1 - 2/5) - 0
= 27/20.