y = 4 + root x

y = (12+x)/3

so, here we have the eq's solved for x

also, I solved for y and got x = (y-4)^2 and x = 3y-12

so looking at the graph I figured the rectangles would do better using f(y)

Thus, we have integration limits y^2 -8y +16 -3y +12 which simplifies to

y^2 - 11y + 28 which simplifies to

(y-7)(y-4) giving us a = 4, b = 7 or

if using c and d for variables

c = 4, d = 7

Now, INT from 4 to 7 of y^2 -11y + 28

= [(7^3)/3 - 11(7^2)/2 +28(7)] - [(4^3)/3 - 11(4^2)/2 +28(4)

Now to make sure I have not flipped signs incorrectly we have:

The first part minus the second part which looks like the first part

- (4^3)/3 + 11(4^2)/2 - 28(4)

So when I did this and got common denominators and combined like terms,

I ended up with a negative value for my final answer and I know if nothing else from this forum that you all have taught me that we cannot have a negative value for Area or Volume. Hence, I have done wrong or performed an error in my operations. So, I am asking for some help from the forum.