Question

Evaluate the line integral using the Fundamental Theorem of Line Integrals. Use a computer algebra system to verify your results. \(\int_{C} e^{x} \sin y d x+e^{x} \cos y d y\) \(C:\) cycloid \(x=\theta-\sin \theta, y=1-\cos \theta\) from (0,0) to \((2 \pi, 0)\)

Short answer

The evaluated line integral is \(e^{2\pi}-1\).

Step-by-step solution

Find the potential function

The given vector field is \(\boldsymbol{F}=e^{x} \sin y \boldsymbol{i}+e^{x} \cos y \boldsymbol{j}\). Find a function \(f(x, y)\) such that \(\partial f / \partial x=e^{x} \sin y\) and \(\partial f / \partial y=e^{x} \cos y\). The anti-derivatives are \(f(x, y)= e^x\sin\,y+k(y)\) and \(f(x, y)=e^x\cos\,y + l(x)\) respectively. Comparing the two equations, we get \(f(x, y)= e^x\sin\,y\) which is the potential function.

Apply the Fundamental Theorem of Line Integrals

The Fundamental Theorem of Line Integrals states that if F is a conservative vector field with potential function f, then for any parametrized curve r(t) with initial point A and terminal point B, \(\int_{C} \boldsymbol{F} \cdot d \boldsymbol{r}=f(B)-f(A)\). Here, A is the initial point (0,0) and B is the terminal point (2π,0). So, \(f(B)-f(A) = e^{(2\pi - \sin 2\pi)}\sin(1-\cos 2\pi) - e^{(0-\sin 0)}\sin(1-\cos 0) = e^{2\pi}\sin1 - e^{0}\sin1 = e^{2\pi} - 1\).

Verify the result using a Computer Algebra System

Using a tool like Mathematica or an equivalent system, run the following command: LineIntegral[{{Exp[x]Sin[y], Exp[x]Cos[y]}, {x,y}}]. This should provide us with a result that matches our calculated result, that is \(e^{2\pi}-1\), confirming our line integral calculation.