Evaluate the integral: ∫c yzdy + xydz, C: x = √t, y = t, z = t^2, ( t= 0 to 1)

no idea how to do this problem, theres normally a ds in the integrals but this one doesnt have one, this is under the chapter "Line Integrals".

Get it all into terms of t

yz * dy + xy * dz =>
t * t^2 * dt + sqrt(t) * t * 2t * dt =>
t^3 * dt + 2t^(5/2) * dt =>
(t^3 + 2t^(5/2)) * dt

Integrate:

(1/4) * t^4 + 2 * (2/7) * t^(7/2) + C =>
(1/4) * t^4 + (4/7) * t^(7/2)

from t = 0 to t = 1

(1/4) * 1 + (4/7) * 1 - (1/4) * 0 - (4/7) * 0 =>
(1/4) + (4/7) =>
(7/28) + (16/28) =>
23/28

Since x = t^(1/2), y = t, z = t^2, we have
dx = (1/2)t^(-1/2), dy = 1 dt, dz = 2t dt.

Therefore, substituting this into the line integral yields
∫c yz dy + xy dz
= ∫(t = 0 to 1) [t * t^2 * 1 dt + t^(1/2) * t * 2t dt]
= ∫(t = 0 to 1) (t^3 + 2t^(5/2)) dt
= [(1/4)t^4 + (4/7)t^(7/2)] {for t = 0 to 1}
= 23/28.

I hope this helps!

all the variable x.y,z should be first converted into t
yzdy+xydz=t*t^2dt+(t^1/2)*t*tdt=(t^3+t^5…
integrating form o to 1 you have
∫(t^3+t^5/2)dt from 0 to 1