Current time:0:00Total duration:6:54

0 energy points

# Stokes example part 4

Video transcript

were now in the home stretch we just have to evaluate the curl of F and then this dot product then we have to evaluate this double integral so the curl of F is going to be equal to- i just remember it as the determinant I J K components, and really you can imagine its the del operator crossed with the actual vector partial with the respect to x- partial with the respect to y partial with the respect to Z and then our vector field it is just equal to negative Y squared is our I component X is our J component and Z squared is our K component and so this is going to be equal to I is going to be equal to I times the partial of Z squared with respect to Y Well there is the Z squared is just a constant with respect to Y so the partial of Z squared with respect to Y is just going to be 0 Minus the partial of X with respect to Z. Once again this is just a constant when you think in terms of Z so thats just going to be 0 nice simplification. Then were going to have minus J- put a negative infront of the J- so we will have The partial of Z squared with respect to X- well thats Zero again and then minus the partial of negative Y squared with respect to Z. Well thats Zero again. and then finally we have our K component So plus K- and K were gonna have the parital of X with respect to X well that actually gives us a value that's just going to be 1 minus the partial of negative Y squared with respect to Y so the partial of negative Y squared with respect to Y is negative 2Y and were subtracting that so its going to be plus 2Y