Please solve the screenshot (handwritten preferred) and explain your work, thanks! Verify that Stokes' Theorem is true for the given vector field F and surface S.F(x, y, z) = -yi + xj – 2k,S is the cone z² = x² + y?, 0 sz5 5, oriented downward.%3D

Concepts introduction: Let S be an oriented smooth surface that is bounded by a simple, closed, smooth boundary curve C with positive orientation. Also let F→ be a vector field then, ∫CF·dr=∬Scurl F ·dS.Given: The vector function Fx,y,z = -yi+xj-2k. The surface S is the cone z2=x2+y2, 0≤z≤5 and oriented downwards.Explanation: To calculate curl F, it is need to calculate the following partial derivatives, P=-y, ∂P∂y=-1, ∂P∂z=0Q=x, ∂Q∂x=1, ∂Q∂z=0R=2, ∂R∂x=0, ∂R∂y=0 Obtain curl F. curl F=∇×F=∂R∂y-∂Q∂zi+∂P∂y-∂R∂zj+∂Q∂y-∂P∂zk=0-0i+0-0j+1--1k=0, 0, 2 Now, determine the normal vector to the cone z2=x2+y2 or z=x2+y2. Since z is nonnegative. n=--∂f∂x, -∂f∂y, 1=∂f∂x, ∂f∂y, -1 When z is 5, the projection on the xy-plane is the circle x2+y2=25, which parameterizes the enclosed region as R=r,θ | 0≤r≤5, 0≤θ≤2π. Obtain the right hand side of the stokes theorem. ∬scurl F·dS=∬s0, 0, 5·∂f∂x, ∂f∂y, -1 dS=∬s-2 dS=-2π52=-50π The intersection of the cone and the plane z=5 is boundary curve C x2+y2=25, z=5, parameterize as rt=5cost, 5sint, 5, r't=-5sint, 5cost, 0, 0≤t≤2π. Obtain the function Frt. Fx,y,z = -yi+xj-2kFrt = ⟨-5sint, 5cost, -2⟩ Now, Frt·r'(t)=-5sint, 5cost, -2·-5sint, 5cost, 0=25sin2t+25cos2t+0=25 Since the surface is oriented downward, calculate the line integral for upward orientation and multiply by negative one. ∫CF·dr=∫02π25dt=25t02π=252π=50π Therefore, ∫CF·dr=∬scurl F·dS.