Using Green's theorum evaluate the integral [ from c x^2ydx-xy^2dy]. Where c is the closed curve along the circle x^2+y^2 = 4

To evaluate the line integral using Green's Theorem, we first need to find the curl of the vector field.

For a vector field F = (P, Q), the curl of F is defined as:

curl(F) = ∂Q/∂x - ∂P/∂y

In this case, P(x, y) = x^2y and Q(x, y) = -xy^2.

Let's compute the partial derivatives:

∂P/∂y = x^2
∂Q/∂x = -y^2

Therefore, the curl of F is:

curl(F) = ∂Q/∂x - ∂P/∂y = -y^2 - x^2

Now, we can find the line integral using Green's Theorem:

∫∫(R) curl(F) dA = ∫∫(R) (-y^2 - x^2) dA

where R is the region enclosed by the given closed curve, which is the circle x^2 + y^2 = 4.

By using polar coordinates, we can rewrite the integral:

∫∫(R) (-y^2 - x^2) dA = ∫∫(R) (-r^2sin^2θ - r^2cos^2θ) r dr dθ

Now, we can evaluate this integral by first integrating with respect to r, and then integrating with respect to θ.

Let's proceed with the calculations.

To evaluate the given integral using Green's theorem, we need to find the line integral along the closed curve C and equate it to the double integral over the region enclosed by C.

Green's theorem states that for a vector field F = (P, Q) and a closed curve C, the line integral of F along C is equal to the double integral of the curl of F over the region enclosed by C. Mathematically, it can be written as:

∮C F · dr = ∬R (∂Q/∂x - ∂P/∂y) dA

In this case, our vector field F = (x^2y, -xy^2). We need to compute the curl of F first:

∂Q/∂x = ∂(-xy^2)/∂x = -y^2
∂P/∂y = ∂(x^2y)/∂y = x^2

So, the curl of F is given by ∂Q/∂x - ∂P/∂y = -y^2 - x^2.

Now, we have to find the region R enclosed by the closed curve C, which is the circle x^2 + y^2 = 4. This region can be expressed in polar coordinates as 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π.

Therefore, the double integral over the region R becomes:

∬R (-y^2 - x^2) dA = ∫₀^(2π) ∫₀² (-r^2) r dr dθ

Integrating with respect to r first:

∫₀^(2π) ∫₀² (-r^2) r dr dθ = ∫₀^(2π) [-r^4/4]₀² dθ
= ∫₀^(2π) -4/4 dθ
= ∫₀^(2π) -1 dθ
= -θ |₀^(2π)
= -2π - 0
= -2π

Thus, the value of the integral ∮C (x^2y dx - xy^2 dy) along the circle x^2 + y^2 = 4 is -2π.