$F(x, y) = x^{10}i + (3x−10{x^9}y)j$
$n$ is the unitary vector normal to curve C
$x²+y²=1$ is the curve C, it is restrict to the first quadrant.
Calculate $\int_{C} \vec F * \vec n ds$
Attempt:
$i = cos(\theta)r - sin(\theta)\theta , j = sin(\theta)r + cos(\theta)\theta$
$\vec F * \vec n = (x^{10}(cos(\theta)\vec r - sin(\theta)\vec\theta) + (3x−10{x^9}y)(sin(\theta)\vec r + cos(\theta)\vec\theta)* \vec r )= x^{10}(cos(\theta)) + (3x−10{x^9}y)sin(\theta)$
$\int_0^{\pi /2} = (x^{10}(cos(\theta)) + (3x−10{x^9}y)sin(\theta)) d \theta = \{cos(\theta)^{10}(cos(\theta)) + (3cos(\theta)−10{cos(\theta)^9}sin(\theta))sin(\theta)\} d \theta$
Now, the answer is -1/11... Unfortunately, the above integration will give 3/2. I am not sure what is wrong.
$$x^2+y^2=1\implies $$ $$x=\cos(t)\; ,y=\sin(t)$$
$$\vec{n}=(\cos(t),\sin(t))$$ $$\vec{F}=(\cos^{10}(t),3\cos(t)-10\cos^9(t)\sin(t))$$ $$ds=dt$$
$$\int_C\vec{F}•\vec{n}ds=$$
$$\int_0^{\frac{\pi}{2}}(\cos^{11}(t)+3\sin(t)\cos(t)-10\cos^9(t)\sin^2(t))dt=$$
$$\Bigl[\cos^{10}(t)\sin(t)-\frac 34\cos(2t)\Bigr]_0^{\frac{\pi}{2}}=$$
$$\frac 32$$