Using Green's theorum evaluate the integral [ integration from c x^2ydx-xy^2dy] where c is the closed curve along the circle. x^2+y^2 = 4
To evaluate the given integral using Green's theorem, follow these steps:
Step 1: Determine the vector field F
The integral is of the form ∮CF · dr, where F = (P, Q) is a vector field. In this case, P = x^2y and Q = -xy^2. Therefore, F = (x^2y, -xy^2).
Step 2: Determine the region enclosed by the closed curve
The curve C represents a circle with radius 2 centered at the origin (0, 0). The region enclosed by this circle is the disk with radius 2.
Step 3: Apply Green's theorem
Green's theorem states that ∮CF · dr = ∬D(∂Q/∂x - ∂P/∂y) dA, where D is the region enclosed by the curve C and dA represents the differential area element.
Let's compute the partial derivatives using P and Q:
∂P/∂y = x^2
∂Q/∂x = -y^2
Step 4: Evaluate the double integral
We need to evaluate ∬D(∂Q/∂x - ∂P/∂y) dA over the disk with radius 2. Since this is a circular region, it is convenient to switch to polar coordinates.
When switching to polar coordinates, we have dA = r dr dθ, and the bounds of integration are r = 0 to 2 and θ = 0 to 2π.
The integrand (∂Q/∂x - ∂P/∂y) becomes:
(-y^2 - x^2)
So, the integral becomes:
∬D(-y^2 - x^2) r dr dθ
Now, integrate in the radial direction first:
∫[0 to 2π] ∫[0 to 2] (-r^2) r dr dθ
Integrating with respect to r gives:
∫[0 to 2π] [-(1/4) r^4] |[0 to 2] dθ
Simplifying:
∫[0 to 2π] [-(1/4) (2^4) - (-(1/4) (0^4))] dθ
= ∫[0 to 2π] [-4] dθ
Integrating with respect to θ gives:
[-4θ] |[0 to 2π]
= -4(2π - 0)
= -8π
Therefore, the value of the given integral using Green's theorem is -8π.