Suppose that a thin metal wire in the shape of a semicircle occupies the curve $C$, where $C$ is the top half of the circle $x^{2}+y^{2}=4$. The mass density of the wire is described by the function $\rho(x, y)=x^{6} y$. Find the total mass of the wire.

Step 1 ：Suppose that a thin metal wire in the shape of a semicircle occupies the curve $C$, where $C$ is the top half of the circle $x^{2}+y^{2}=4$. The mass density of the wire is described by the function $\rho(x, y)=x^{6} y$. We are asked to find the total mass of the wire.

Step 2 ：The total mass of the wire can be calculated by integrating the mass density function over the curve $C$. Since the curve is a semicircle, we can parameterize it using polar coordinates.

Step 3 ：We can let $x = r\cos(\theta)$ and $y = r\sin(\theta)$, where $r = 2$ (the radius of the circle) and $\theta$ ranges from $0$ to $\pi$. Then, the mass density function becomes $\rho(r, \theta) = (2\cos(\theta))^{6} \cdot 2\sin(\theta)$.

Step 4 ：The total mass is then the integral of this function with respect to $\theta$ from $0$ to $\pi$.

Step 5 ：By calculating the integral, we find that the total mass of the wire is $\frac{256}{7}$.

Step 6 ：Final Answer: The total mass of the wire is \(\boxed{\frac{256}{7}}\).