Question

Evaluate using Green's Theorem: https://i.imgur.com/vanoNC6.png

Answer 1

∳Pdx+Qdy=∬(∂Q/∂x-∂P/∂y)dA is the general form.

In this problem P=x²y, Q=x, ∂Q/∂x=1,∂P/∂y=x².

So we have:

∬(1-x²)dA over the given region.

Triangular closed region has base BC=line segment (0,0)→(1,0).

Vertical leg is segment C(1,0)→A(1,2). The double integral is anti-clockwise B→C→A→B (loop encloses the region).

dA=dxdy or dydx.

Inner integral ∫(1-x²)dy[0,2x]=[y-x²y]²ˣ₀=2x-2x³.

Outer integral ∫(2x-2x³)dx[0,1]=[x²-x⁴/2]¹₀=1-½=½.