Question

Let P(x, y)=3 y-x² y, Q(x, y)=2 x+x y. Let C be the unit circle center at (0,0) with clockwise orientation, and let

D={(x, y): x²+y²≤ 1} be the unit disk with boundary ∂D.
Complete the Green's theorem: ∫_∂D (3y−x²y)dx+(2x+xy)dy=___

Answer 1

By applying Green's theorem, the given vector field's circulation on the unit disk's boundary is computed by evaluating the double integral of the divergence of the field over the disk's area, using partial derivatives and polar coordinates.

Step-by-step explanation:

To apply Green's theorem to a given vector field over a certain region, we need to consider the partial derivatives of the functions that make up the field. The given vector field is P(x, y) = 3y - x²y and Q(x, y) = 2x + xy. According to Green's theorem, the circulation of this vector field around the boundary of the region, in this case, the unit disk D, is equal to the double integral over the region D of the partial derivative of Q with respect to x minus the partial derivative of P with respect to y.

Firstly, calculate the partial derivatives:

1. ∂Q/∂x = 2 + y
2. ∂P/∂y = 3 - x²
   Then, the circulation around ∂D (boundary of D) is given by the integral over D:

∬_D (∂Q/∂x - ∂P/∂y) dA = ∬_D (2 + y - (3 - x²)) dA = ∬_D (x² + y - 1) dA.

Since D is the unit circle, we can set up the double integral in polar coordinates where x = rcos(θ) and y = rsin(θ), and the area element dA is r dr dθ with r ranging from 0 to 1 and θ ranging from 0 to 2π:

∬_D (rcos(θ)² + rsin(θ) - 1) r dr dθ = ∬_D (r³cos(θ)² + r²sin(θ) - r) dr dθ.

Now, we evaluate the integral using the symmetry of the unit circle and known integrals of trigonometric functions and powers of r:

∬_0^{2π} ∬_0^1 (r³cos(θ)² + r²sin(θ) - r) dr dθ.