With plane x+ 2y+ 2z= 4, we can write x= 4- 2y- 2z and so a position vector for a point on the plane is given by . The two derivatives, and . The differential, in terms of the parameters y, z, is given by their cross product: . Of course, I have chosen the order of vectors in of the product to give a positive component "oriented with outward normal point upward".
Take the dot product of that with the given vector, and integrate. Projecting into the yz- plane, 2y+ 2z= 4 so y goes from 0 to 2 and, for each y, z goes from 0 to 2- y.