I have question about integration ,I know that's not difficult.I stuck on this ,As honestly as possible I forget this section.
$$\int_{c} (3a(x^2+y^2)-y^3)dx+3xy(2a-y)dy =?$$ and 'c' is higher part of this cirlce with horizontal line like below $$c: y=a+\sqrt{a^2-x^2}$$
Thanks in advanced for any hint .
Hint:
For the straight part, $-a\le x\le a$ and $y=a$ (and $dy=0$) so that you integrate
$$\int_{x=-a}^a(3a(x^2+a^2)-a^3)dx.$$
For the arc,
$$\int_{x=a}^{-a} (3a(x^2+(a+\sqrt{a^2-x^2})^2)-(a+\sqrt{a^2-x^2})^3)dx+3x(a+\sqrt{a^2-x^2})(2a-(a+\sqrt{a^2-x^2}))\frac{-x\,dx}{\sqrt{a^2-x^2}}.$$
Simplifications occur, and probably a trigonometric substitution such as $x=a\cos t$ will help.
We can see the vector field is conservative, since
$$\frac{\partial}{\partial x}\left(3xy(2a-y)\right)=\frac{\partial}{\partial y}\left(3a(x^2+y^2)-y^3\right)$$
and thus the integral over the whole closed path is zero, and thus you'd have only to calculate the value over the straight line and substract this:
$$C: r(t):=(t,a)\;,\;\;-a\le t\le a\implies r'(t)=(1,0)\implies$$
$$F(r(t))=\left(\,3a(t^2+a^2)-a^3,\,3a^2t\,\right)\implies F(r(t))\cdot r'(t)=3at^2+2a^3\implies$$
$$\int_C\vec F\cdot d\vec r=\int_{-a}^a(3at^2+2a^3)dt=2a^4+2a^4\;\ldots$$