Finding a line integral of $F(x,y) = (3x^2\cos y + 2\cos x, -x^3\sin y)$ along a given curve.

Question

Let $F$ : $R^2 \to R^2$ be the vector field F(x,y) = ($3x^2cosy+2cosx,-x^3siny$) and $\gamma$ : [$0$,$\pi$]$\to$$R^2$ be the curve $\gamma(t)=(t,(\pi-t)^2)$. Find the line integral of $F$ along $γ$ ?
I tried to solve the problem: Firstly I found F($\gamma(t)$) = ($3t^2cos(\pi-t)^2+2cost$,$-t^3sin(\pi-t)^2$).
Secondly I found the derivative of $\gamma'(t)$ = (1,-2($\pi$-t)) The next thing which I have to do is finding the dot product of F($\gamma(t)$) and $\gamma'(t)$. I found it but I could not take integral from it.

Answer 1

Hint. We first need the vector field evaluated along the given curve: $$ \vec{F}(\vec{\gamma}(t))=(3t^2\cos((\pi-t)^2) + 2\cos t, -t^3\sin((\pi-t)^2)) $$ next we need the derivative of the parameterization $$ \vec{\gamma'(t)}=(1,-2(\pi-t)). $$ Finally, let’s get the dot product $$ \begin{align} \vec{F}(\vec{\gamma}(t))\cdot \vec{\gamma'(t)}&=(3t^2\cos((\pi-t)^2)+2\cos t)\cdot 1+(-t^3\sin((\pi-t)^2))\cdot (-2(\pi-t)) \\\\&=3t^2\cos((\pi-t)^2)+2\cos t+2t^3(\pi-t)\sin((\pi-t)^2)) \end{align} $$ then we have to evaluate $$ \int_{C}\vec{F}\cdot d\vec{\gamma}=\int_0^{\pi}\vec{F}(\vec{\gamma}(t))\cdot \vec{\gamma'(t)}\:dt. $$ Can you take it from here?

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Edit. One may integrate, just observing that $$ (t^3\cos((\pi-t)^2))'=3t^2\cos((\pi-t)^2)+2t^3(\pi-t)\sin((\pi-t)^2). $$

Answer 2

Look the path is from $\;(0,\pi^2)\;$ to $\;(\pi,0)\;$ , and the vector field is conservative because

$$\frac{\partial}{\partial y}(3x^2\cos y+2\cos x)=-3x^2\sin y=\frac{\partial}{\partial x}(-x^3\sin y)$$

Then we can go between those two points as we want and not necessarily on $\;\gamma\;$, so I choose the straight lines from $\;(0,\pi^2)\;$ to origin, and then from origin to $\;(0,\pi)\;$ :

$$r_1(t):=t(0,0)+(1-t)(0,\pi^2)=\left(0\,,\,\,\pi^2(1-t)\right)\,,\;\;0\le t\le 1$$ and then

$$r_2(t):=t(\pi,0)+(1-t)(0,0)=(\pi t,0)\,,\,\,0\le t\le 1$$ and the line integrals are

$$\int_0^1 F(r_1(t))r_1'(t)dt=\int_0^1\left(0,0\right)\left(0,\,-\pi^2\right)dt=\int_0^10\,dt=0$$

$$\int_0^1F(r_2(t))r_2'(t)dt=\int_0^1\left(3\pi^2t^2+2\cos\pi t,\,0\right)\left(\pi,0\right)dt=$$

$$=3\pi^3\int_0^1 t^2dt+2\pi\int_0^1\cos\pi t\,dt=\pi^3$$

Answer 3

A different approach is to notice that $F$ is conservative. You can thus instead compute the integral either by choosing a more convenient path, as in this answer, or you can find a scalar potential $\phi$ such that $F=\nabla\phi$ and evaluate $\phi(g(\pi))-\phi(g(0))$.

There are a couple of ways to find $\phi$. One method is to alternately integrate and differentiate the components of $F$ with respect to its parameters: $$\phi(x,y) = \int 3x^2\cos y+2\cos x\,dx = 2\sin x+x^3\cos y+h(y)$$ and ${\partial\phi\over\partial y}=-x^3\sin y+h'(y) = -x^3\sin y$, the $y$-component of $F$, which means that $h=\text{const}$. Since we’ll be subtracting two values of $\phi$ from each other, we can take $h$ to be zero for simplicity.

Alternatively, since $F$ is defined in a star-shaped region relative to the origin, we can directly compute $$\phi(x,y)=\int_0^1F(tx,ty)\cdot(x,y)\,dt,$$ which is just the line integral of $F$ along the segment joining the origin to the point $(x,y)$. For this problem, the first method looks simpler.