Calculate the integral over the boundary of d of the function capital F dotted with dr where D is the annulus given by the polar inequalities, 2 is less than or equal to r is less than or equal to 5 and 0 is less than or equal to theta is less than or equal to 2 pi and capital F of x equals angle bracket x cubed comma 5x plus e to the y end exponent times sine y close angle bracket. Go ahead and appeal to Green's theorem once again and find qx minus py.

Just be the double integral along the boundary of d of Q sub x minus P sub y, and that is with respect to the area, dA. So Qx would be equal to 5, and Py is equal to, Q sub x is equal to 5, P sub y equals 0. So this becomes the double integral or the integral. Yeah, double integral over the boundary of D of 5, dA, because this is a constant this is equal to 5 times the area of the annulus.

The area of the annulus is the outer circle, area of the outer circle, minus the area of the inner circle. So this would be 5 parentheses 25 pi minus 4 pi, going back to our radius values of 2 and 5 respectively. So that this would be equal to 105 pi.