Calculus III - Fundamental Theorem for Line Integrals (Practice Problems)

Hide Mobile Notice

Mobile Notice

You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best viewed in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (you should be able to scroll/swipe to see them) and some of the menu items will be cut off due to the narrow screen width.

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.
Solution