Section 16.5 : Fundamental Theorem for Line Integrals
Evaluate \( \displaystyle \int\limits_{C}{{\nabla f\centerdot d\vec r}}\) where \(f\left( {x,y} \right) = {x^3}\left( {3 - {y^2}} \right) + 4y\) and C is given by \(\vec r\left( t \right) = \left\langle {3 - {t^2},5 - t} \right\rangle \) with \( - 2 \le t \le 3\). Solution
Evaluate \( \displaystyle \int\limits_{C}{{\nabla f\centerdot d\vec r}}\) where \(f\left( {x,y} \right) = y{{\bf{e}}^{{x^{\,2}} - 1}} + 4x\sqrt y \) and C is given by \(\vec r\left( t \right) = \left\langle {1 - t,2{t^2} - 2t} \right\rangle \) with \(0 \le t \le 2\). Solution
Given that \( \displaystyle \int\limits_{C}{{\vec F\centerdot d\vec r}}\) is independent of path compute \( \displaystyle \int\limits_{C}{{\vec F\centerdot d\vec r}}\) where C is the ellipse given by \(\displaystyle \frac{{{{\left( {x - 5} \right)}^2}}}{4} + \frac{{{y^2}}}{9} = 1\) with the counter clockwise rotation. Solution
Evaluate \( \displaystyle \int\limits_{C}{{\nabla f\centerdot d\vec r}}\) where \(f\left( {x,y} \right) = {{\bf{e}}^{x\,y}} - {x^2} + {y^3}\) and C is the curve shown below.
