What is the Result of Evaluating a Line Integral Along a Quarter-Circle Curve?

[DryRun]

Homework Statement
Evaluate $\int3x^2dx+2xydy$, where C is the curve $x^2+y^2=4$ starting at (2,0) and ending at (0,2) in the anti-clockwise direction.

The attempt at a solution
The curve C is a quarter-circle with centre (0,0) and radius=2.
Making y subject of formula:
$$y=+\sqrt{4-x^2}$$ since the quarter-circle is above the x-axis.
$$\frac{dy}{dx}=-\frac{x}{\sqrt{4-x^2}}$$
Replacing y and dy in the line integral.
$$\int3x^2dx+2xydy=\int_{x=2}^{x=0} 3x^2dx+2x.\sqrt{4-x^2}.-\frac{x}{\sqrt{4-x^2}}.dx=\int_{x=2}^{x=0} x^2dx=\frac{x^3}{3}\Biggr|_2^0=(0-\frac{8}{3})=-\frac{8}{3}$$
Is this correct?

 

[DryRun]

My friend told me that this problem should be solved by converting to polar coordinates. But i did it directly without any conversion, which is why I'm wondering if it's good at all?

 

[ehild]

It is correct.

ehild

 

[DryRun]

Thank you for your confirmation, ehild.