Given vector field:H =  𝑥2𝑦2 ay — 𝑥2𝑒−𝑦ax Solve both sides of the STOKES Theorem with the given vector field and that passes through the path with vertices:P1 (3, 2)P2 (1, -3)P3 (-4, 2)

Stokes' theorem states that the integral of a vector field over a surface is equal to the integral of the curl of the field over the boundary of the surface.

Given the vector field H = x²y² ay - x²e^-y ax, we first need to compute the curl of H.

The curl of a vector field F = P(x,y,z) i + Q(x,y,z) j + R(x,y,z) k is given by:

curl F = (∂R/∂y - ∂Q/∂z) i - (∂R/∂x - ∂P/∂z) j + (∂Q/∂x - ∂P/∂y) k

In this case, our vector field is in two dimensions, so we can consider it as H = -x²e^-y i + x²y² j + 0 k.

Then, curl H = (∂/∂y (0) - ∂/∂z (x²y²)) i - (∂/∂x (0) - ∂/∂z (-x²e^-y)) j + (∂/∂x (x²y²) - ∂/∂y (-x²e^-y)) k

Solving these partial derivatives, we get curl H = 0 i - 0 j + (2x²y - x²e^-y) k

Now, we need to compute the line integral of H over the boundary of the surface. The boundary is the triangle with vertices P1 (3, 2), P2 (1, -3), and P3 (-4, 2).

We parameterize this path as r(t) = (1-t) P1 + t P2 for 0 <= t <= 1 (from P1 to P2), r(t) = (2-t) P2 + (t-1) P3 for 1 <= t <= 2 (from P2 to P3), and r(t) = (3-t) P3 + (t-2) P1 for 2 <= t <= 3 (from P3 to P1).

Then, we compute the line integral ∫ H . dr over each segment and add them up. This involves substituting the parameterization into H, computing the derivative dr/dt, and integrating from the start to the end of each segment.

Finally, we compare the result of the line integral with the double integral of curl H over the surface. The surface integral is ∫∫ (curl H) . dS, where dS is the differential area element. In this case, dS = dx dy k, so we are integrating (2x²y - x²e^-y) dx dy over the triangle.

If Stokes' theorem holds, these two results should be equal.