# Calculating the work integral of a vector field.

Let F be the vector field

F = $<\dfrac{1}{x} + 2x + y, \ \ \dfrac{1}{y} + x + 1>$ .

Compute the work integral $\int_{c} F dr$ where C is the path

r(t) = (1 + cos t)i + (2 + sin t) j $\ \ \ \ -\dfrac{\pi}{2} \leq t \leq \dfrac{\pi}{2}$.

I am very confused as to how to approach this problem. I understand that we simply have to parameterize F, find v(t), and dot product both of them, but given these expressions that would be very tedious. Is there a simpler way to approach this problem?

\eqalign{ & W = \int\limits_C {F \bullet dr} \cr & - - - - Test\;For\;Gradient\;Field \cr & F = \left( {{1 \over x} + 2x + y\,,\,{1 \over y} + x + 1} \right) = \left( {M\;,\;N} \right) \cr & F = \nabla f\;\;if\;\;\;{\partial _y}M = {\partial _x}N \cr & {\partial _y}M = 1 \cr & {\partial _x}N = 1 \cr & W = \int\limits_C {F \bullet dr} = \int\limits_C {\nabla f \bullet dr} = f({p_1}) - f({p_0}) \cr & \nabla f = \left( {{1 \over x} + 2x + y\,,\,{1 \over y} + x + 1} \right) = \left( {{\partial _x}f\;,\;{\partial _y}f} \right) \cr & - - - - Find\;\;f(x,y) \cr & \left\{ \matrix{ {\partial _x}f = {1 \over x} + 2x + y \hfill \cr {\partial _y}f = {1 \over y} + x + 1 \hfill \cr} \right. \cr & \int {{\partial _x}fdx = \ln (x) + {x^2}} + yx + g(y) = f(x,y) \cr & {\partial _y}f = x + g'(y) \cr & g'(y) = {1 \over y} + 1 \cr & g(y) = \ln (y) + y + c \cr & f(x,y) = \ln (x) + {x^2} + yx + \ln (y) + y + c \cr & - - - - Find\;\;{p_1}\;and\;{p_0} \cr & x = 1 + \cos t \cr & y = 2 + \sin t \cr & {{ - \pi } \over 2} \le t \le {\pi \over 2} \cr & 1 \le x \le 1 \cr & 1 \le y \le 3 \cr & {p_0} = (1,1) \cr & {p_1} = (1,3) \cr & - - - - Finding\;The\;Work \cr & W = \int\limits_C {F \bullet dr} = \int\limits_C {\nabla f \bullet dr} = f({p_1}) - f({p_0}) \cr & W = \left( {\ln (1) + {1^2} + 3 + \ln (3) + 3} \right) - \left( {\ln (1) + {1^2} + 1 + \ln (1) + 1} \right) = \ln (3) + 4 \cr}