Find the volume of the solid under the surface z = 5x + 5y^2 and above the region bounded by x = y^2 and x = y^3

To solve this problem, we have to do derivation and integration. I will just describe to you the steps and do the calculation on paper.

x = y^2 = y^3 gives (x,y) = (0,0) and (1,1)

Therefore the base of the solid is the area between the two curves bounded by the origin and point (1,1)

The differential equation based on the given is:
dV = [5x + 3*y^2]dx*dy 
dV = [5x*dx]dy + [3*y^2*dy]dx 

The 1st thing to do is to integrate the first term in x and the bounds are from y^3 to y^2. Then integrate over y and the bounds are from y = 0 to y = 1. 

Then integrate the second term over y and evaluate from y = x^(1/2) toy = x^(1/3). Then integrate over x and evaluate from x = 0 to x = 1. 

First term: 
Integrate x and evaluate: (5/2)x^2 = (5/2)[y^4 - y^6] 
Integrate y: (5/2)[y^5/5 - y^7/7] 
Evaluate: (5/2){[1/5 - 1/7] - [0]} = 5/70 

Second term: 
Integrate y and evaluate: y^3 = x - x^(3/2) 
Integrate x: x^2/2 - (2/5)x^(5/2) 
Evaluate: [1/2 - 2/5] - [0] = 1/10 

Volume = 5/70 + 1/10 = 12/70 
Volume = 0.1714           (ANSWER)

its line 2 i just added the rest or else it wouldn't let me answer