Help with evaluating a line integral

Could someone please help me with this problem? I am so stuck on it. I was attempting to do in 3 parts, but I think my parameterizing (and especially my interval for each integral) is incorrect. I did it using Green's theorem and got -16, which is definitely not what I got in my attempt here. I would love to be walked through this! Especially the setting up. Thank you so very much!

Line integral ∫ (y^2 dx + 6xy dy). (with a C at the bottom of the integral, but couldn't figure out how to put that in).
C is the boundary of the region bounded by y=sqrt(x), y=0 and x=4. Oriented in the clockwise direction.

Evaluate line integral directly by parameterizing C.

Let C1 be the curve x = t, y = sqrt(t), for 0 <= t <= 4.
dx = dt, dy = 1/(2sqrt(t)) dt

∫C1 of (y^2 dx + 6xy dy)
= ∫0 to 4 of (t dt + 6tsqrt(t)/(2sqrt(t)) dt)
= ∫0 to 4 of (4t dt)
= 2*4^2 - 2*0^2 = 32.

Let C2 be the segment x = 4, y = 2 - t, for 0 <= t <= 2.
dx = 0 dt, dy = -dt

∫C2 of (y^2 dx + 6xy dy)
= ∫0 to 2 of (0 dt + 6(4)(2 - t)(-dt))
= ∫0 to 2 of (24(t - 2)dt)
= 12(2 - 2)^2 - 12(0 - 2)^2
= -48

Let C3 be the segment x = 4 - t, y = 0, for 0 <= t <= 4.
dx = -dt, dy = 0 dt

∫C3 of (y^2 dx + 6xy dy)
= ∫0 to 4 of (0 dt + 0 dt)
= 0

Finally,
∫C of (y^2 dx + 6xy dy)
= ∫C1 of (y^2 dx + 6xy dy)+∫C2 of (y^2 dx + 6xy dy)+∫C3 of (y^2 dx + 6xy dy)
= 32 - 48 + 0
= -16, which matches the answer that you got by using Green's Theorem