Were given a vector field F and a curve that is oriented clockwise is viewed from above sea and were asked to Stokes his serum to evaluate the line. Integral oversee of F so f is thieve vector field one I plus X plus y zj plus X y minus root C, K and C is the boundary of the part of the plane. Three X plus two y plus C equals one in the first often. So first, let's calculate the curl of F. This is equal to X minus. Why I and then we have minus why j plus okay. And we have their surface s. You were told. This is bounded by the curve which bounds the portion of the plane. Three X plus two y plus Z equals one over. We're in the first often so the domaine de, which is the set of pairs X y such that being the first often X has to lie between zero and one third. And then why lies between zero and the line one half of one minus three x, and we want to orient s upward, since our curve is counterclockwise when viewed from above, and we have their surface can also be described as a function of X and Y. So this will simplify her calculations a little bit. So Z equals G of X y, which equals one minus three X minus two y and therefore have that the line integral oversee of F This is by Stokes is there equal to the surface integral over s of the curl of F and evaluating This is the double integral over the domaine de of and here we're using the shortcut that I mentioned before we have the opposite of the X component of Sorry, the first component of our vector field f So the opposite of X minus y times the partial derivative of Z prostrated of G, if you like with respect to X which is negative three minus the second component of F, which is X plus y z times part of derivative of G. With respect to why which is negative two plus the third component of F, which is X Y minus root C d. A. And plugging in Z is in terms of x and y. This reduces to the integral from writing is an iterated integral from Mexico. 01 3rd integral from 0 to 1 half of three x minus one of simplifies to one plus three X minus five y All right, de y dx actually in the steak here. Sorry. This should be the components of the curl of F instead instead of the components of F. So this is instead of X plus y Z. This is negative y times negative too. And instead of x y minus root, see, this is simply one makes a little more sense. So we obtained this and then taking the anti derivative with respect to why we get integral from 01 3rd of why plus three x y minus five hats y squared from what I equals zero toe y equals one half of three X minus one The X and substitution gives the integral from zero toe one third. Um, and here we have one half times one plus three x Times three x minus one minus five halves times three x minus one squared times one half squared. So if I've had times of fourth times three x minus one squared DX and multiplying the south. We get the integral from zero to one third of Let's see, we have X squared term is negative. 81 8th x squared The ex term is 15 4th x, and the constant term is negative. 1/8 DX Taking the anti derivative We get negative 27 8th x Q plus 15 8th X squared minus 1/8 x from zero toe, one third Plugging in. We get negative 1/8 plus five 24th minus one 24th and this simplifies to one 24th.