Line integral along the curve $\gamma(t)=(4 \cos t, 4 \sin^2 t) $

Question

Let us consider the vector field in the plane: $$\vec{F}=\left(x \frac{e^{x^2+y^2} - e}{x^2+y^2},y\frac{e^{x^2+y^2} - e}{x^2+y^2}\right)$$ calculate the line integral along the curve defined by: $$\gamma: \begin{cases} x=4 \cos t \\ y=4 \sin^2 t\\ \end{cases} $$ with $t\in[0,\pi/2]$.

Any suggestions please?

Answer 1

As Git Gud remarked, the integral is independent of path (because the vector field F is conservative). So choose the path along (4cos(t), 4sin(t)) instead. The answer is eventually 0.

Answer 2

At first compute $\vec F(t)$ by substituting the expressions $x,y$ given by your curve $\gamma$. Then compute the differentials $dx,dy$ for $x,y \in \gamma$ in dependence on $t$. Finally, evaluate the scalar product $(dx,dy)\vec F$ and calculate the integral over the interval of $t$.

Answer 3

Here is a direct evaluation using this link

$\begin{align} dx&=-4\sin t \\ dy&=8\cos t \sin t \\ F&=\Big(\frac{2[\exp{(2\cos 4t +14)}-e] \cos t }{7+\cos 4t},\frac{2[\exp{(2\cos 4t +14})-e] \sin^2 t }{7+\cos 4t}\Big) \end{align}$

Your integral will hence be

$\begin{align}I&=-\int_0^{\frac{\pi}{2}}\frac{2[\exp{(2\cos 4t +14})-e] \sin 4t }{7+\cos 4t}dt\\ &=-\int_0^{\frac{\pi}{4}}\frac{2[\exp{(2\cos 4t +14})-e] \sin 4t }{7+\cos 4t}dt-\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{2[\exp{(2\cos 4t +14})-e] \sin 4t }{7+\cos 4t}dt\\ &=-\int_0^{\frac{\pi}{4}}\frac{2[\exp{(2\cos 4t +14})-e] \sin 4t }{7+\cos 4t}dt+\int_0^{\frac{\pi}{4}}\frac{2[\exp{(2\cos 4t +14})-e] \sin 4t }{7+\cos 4t}dt\\ &=0. \end{align}$

where at last I have changed the variable for the second integral from $t$ to $\frac{\pi}{2}-t$.