Given $\oint_c \tan^{-1}(x)dx + 3xdy $ and C is a rectangle with vertices $(1,0),(0,1),(2,3),(3,2).$ Here is what I have so far : $$\oint_c \tan^{-1}(x)dx + 3xdy = \int\int_R (\frac{dn}{dx}-\frac{dm}{dy})$$ $$ M(x,y) = \tan^{-1}(x), \space N(x,y)=3x$$ $$\oint_c \tan^{-1}(x)dx + 3xdy = \int\int_R\ (\frac{d}{dx}(3x) - \frac{d}{dy}\tan^{-1}(x) dydx ) $$ $$= \int_0^3\int_0^3 (3 - 0)dydx$$ $$=\int_0^3 3y \vert_0^3 \space dx$$ $$=\int_0^3 9 \space dx$$ $$=\int_0^3 9x \vert_0^3 \space dx$$ $$= 27$$
How would I sketch this and is this the correct way to evaluate this line integral using Green's theroem ?
The region is $[0,3] \times [0,3]$, the region that you used in your integration.