HINTS: The parametrisation of a circle of radius $r$ in $\mathbb{R}^2$ is given by
$$\gamma(t) = (r \cos (t), r \sin (t)).$$
Arclength is calculated by working out
$$\int_a^b | \gamma'(t)| dt,$$
where $a, b$ are the two ends of your interval and $|\gamma ' (t)|$ is the velocity of the cure.
EDIT: Ok, assuming you have made some kind of attempt, here is how to answer the question:
We are told that the radius of the circle is $3$, so using my first hint, we can write the circle as
$$\gamma(t) = (3 \cos (t), 3 \sin (t)).$$
Now, we want to calculate the arc length of the upper half. We know that a regular circle goes from $0$ round to $2 \pi$ and so if we want just the upper half, we take half of this and we get our interval to be from $0$ and $\pi$, i.e we have $a = 0$ and $b = \pi$.
Next, we want to calculate $| \gamma ' (t)|$:
$$\gamma(t) = (3 \cos (t), 3 \sin (t)),$$
$$\gamma'(t) = (-3 \sin (t), 3 \cos (t)),$$
$$|\gamma'(t)| = |(-3 \sin (t), 3 \cos (t)| = \sqrt{(-3 \sin (t))^2 + (3\cos (t))^2} = \sqrt{9\sin ^2(t) + 9 \cos ^2(t)} = \sqrt{9(\cos ^2(t) + \sin ^2(t))} = 3,$$
which means to find the arc length, we now have to solve the equation
$$\int_{0}^{\pi} 3 dt$$
$$= [3t]_{0}^{\pi}$$
$$= 3(\pi) - 3(0) = 3\pi.$$
Now, if you wanted to check to see if your answer is correct, you can use a different method, i.e like one Andre Nicolas used, and you will see that you get the same answer.
Just a word of the wise: I don't know what went through your head, but remember that your final equation should be equal to the absolute value of the previous. This means you should probably be looking for points of intersection with the y-axis, but you might have just seen the equation and thought "Oh, this is a fourth degree equation, so the result will always be positive," but I can't really tell.
Other than that, if you just evaluated the integral at 4 and 2 and subtracted the two numbers, you should be golden :)
Best Answer
Hint: This problem is much easier tackled viewing $x$ as a function of $y$ instead of the other way around. We have
$$x=f(y)=\sqrt{12}y^{3/2},\\ f'(y)=3\sqrt{3y},\\ [f'(y)]^2=27y.$$
Can you take it from there?