Let F (x, y) = (e^{−y} − 2x, −xe^{−y} − sin(y)) = (M(x, y), N(x, y)) Find ∫_{C}(e^{−y}−2x)dx+(−xe^{−y}−sin(y))dy, where C is a piecewise C^{1} curve starting at (2, 0) and ending at (0, 0). Does \frac{∂N}{∂x} = \frac{∂M}{∂y}? \frac{∂N}{∂x} = \frac{∂(−xe^{−y}−sin(y))}{∂x} = e^{−y} \frac{∂M}{∂y} = \frac{∂(e^{−y}−2x)}{∂y} = e^{−y} Thus \frac{∂N}{∂x} = \frac{∂M}{∂y}. Thus F is a gradient field and hence has path independent integrals. F = ∇f = (\frac{∂f}{∂x}, \frac{∂f}{∂y}) = (M, N) = (e^{−y} − 2x, −xe^{−y} − sin(y)) Thus \frac{∂f}{∂x} = e^{−y} − 2x Hence f(x, y) = ∫ (e^{−y} − 2x)dx = xe^{−y} − x^{2} + c(y). Thus \frac{∂f}{∂y} = \frac{∂(xe^{−y}−x^{2}+c(y))}{∂y} = −xe^{−y} + c′(y) Thus −xe^{−y} − sin(y) = \frac{∂f}{∂y} = −xe^{−y} + c′(y). Thus c′(y) = −sin(y) and c(y) = ∫ (−sin(y))dy = cos(y) + k Hence f(x, y) = xe^{−y} − x^{2} + cos(y) + k Since F is a gradient field, ∫_{C}(e^{−y} − 2x)dx + (−xe^{−y} − sin(y))dy = f(0, 0) − f(2, 0) = 0 − 0 + 1 + k − [2 − 4 + 1 + k] = 2 Alternatively, let X : [0, 2] → R^{2}, X(t) = (2 − t, 0) = (x(t), y(t)). Then X(0) = (2, 0) and X(2) = (0, 0)

x(t) = 2 − t implies dx = −dt

y(t) = 0 implies dy = 0.

∫_{C}(e^{−y} − 2x)dx + (−xe^{−y} − sin(y))dy = ∫_{0}^{2}(1 − 2(2 − t))(−dt) = ∫_{0}^{2}(3 − 2t)dt = 3t − t^{2}|_{0}^{2} = 6 − 4 = 2.

Find ∫_{C}(e^{−y}−2x)dx+(−xe^{−y}−sin(y))dy, where C is a piecewise C^{1} curve starting at (2, 0) and ending at (2, 0).

Scalar line integral over a curve X : [a, b] → R^{n}, f : R^{n} → R: ∫_{X} fds = ∫_{a}^{b} f(x(u))||x′(u)||dt = ∫_{a}^{b} f(x(u))||t(u)||dt

Vector line integral over a curve X : [a, b] → R^{n}, F : R^{n} → R^{n}: ∫_{X} F · ds = ∫_{a}^{b} F (x(u)) · x′(u)dt = ∫_{a}^{b} F (x(u)) · t(u)dt = ∫_{a}^{b} F (x(u))· \frac{t(u)}{||t(u)||}||t(u)|| = ∫_{a}^{b} F (x(u))·T (u)||t(u)||dt = ∫_{a}^{b} F ·T ds where T (u) is the unit tangent to the curve x at u.

Scalar surface integral over a surface X : D ⊂ R^{2} → R^{3}, f : R^{3} → R: ∫ ∫_{X} fds = ∫ ∫_{X} f(x(u, v))||N(u, v)||dudv

Vector surface integral over a surface X : D ⊂ R^{2} → R^{3}, F : R^{3} → R^{3}: ∫ ∫_{X} F · ds = ∫ ∫_{X} F (x(u, v)) · N(u, v)dudv = ∫ ∫_{X} F (x(u, v)) · \frac{N(u,v)}{||N(u,v)||}||N(u, v)||dudv = ∫ ∫_{X} F (x(u, v)) · n(u, v)||N(u, v)||dudv = ∫_{a}^{b} F · n ds where n(u) is the unit normal to the surface S at u.