Evaluate integral_C(x2 + y2) ds, where C is a circle of radius R = 4 centered at the origin.

The curve $C$ is a circle of radius $R=4$ centered at the origin. We can parameterize it using $t$ as the angle:

$$x(t) = 4 \cos t$$ $$y(t) = 4 \sin t$$

for $0 \le t \le 2\pi$.
So, the position vector is $r(t) = \langle 4 \cos t, 4 \sin t \rangle$.

First, find the derivative of the parameterization:

$$r'(t) = \left\langle \frac{d}{dt}(4 \cos t), \frac{d}{dt}(4 \sin t) \right\rangle = \langle -4 \sin t, 4 \cos t \rangle$$

Next, find the magnitude of $r'(t)$:

$$|r'(t)| = \sqrt{(-4 \sin t)^2 + (4 \cos t)^2}$$ $$|r'(t)| = \sqrt{16 \sin^2 t + 16 \cos^2 t}$$ $$|r'(t)| = \sqrt{16(\sin^2 t + \cos^2 t)}$$

Using the identity $\sin^2 t + \cos^2 t = 1$:

$$|r'(t)| = \sqrt{16(1)} = \sqrt{16} = 4$$

So, $ds = |r'(t)| \, dt = 4 \, dt$.

Substitute $x=4 \cos t$ and $y=4 \sin t$ into the integrand:

$$f(x(t), y(t)) = (4 \cos t)^2 + (4 \sin t)^2$$ $$f(x(t), y(t)) = 16 \cos^2 t + 16 \sin^2 t$$ $$f(x(t), y(t)) = 16(\cos^2 t + \sin^2 t)$$

Using the identity $\sin^2 t + \cos^2 t = 1$:

$$f(x(t), y(t)) = 16(1) = 16$$

The line integral is given by $\int_C f(x,y) \, ds = \int_a^b f(x(t), y(t)) |r'(t)| \, dt$.
Substitute the expressions found in the previous steps:

$$\int_C (x^2 + y^2) \, ds = \int_0^{2\pi} (16)(4) \, dt$$ $$\int_0^{2\pi} 64 \, dt$$

Now, evaluate the definite integral:

$$[64t]_0^{2\pi} = 64(2\pi) - 64(0)$$ $$= 128\pi$$

The value of the integral is $128\pi$.