Answer
$$1638.4$$
Work Step by Step
1. According to equation 3:
$$\int_C f(x,y)ds = \int_a^bf(x(t),y(t))\sqrt {(\frac {dx}{dt})^2 + (\frac{dy}{dt})^2}dt$$
According to the equation: $x^2 + y^2 = 16$, this circle has a radius of $\sqrt {16} = 4$ and it is centered at (0,0). Thus, the parametric equations are: $x = 4cos(t)$ and $y = 4sin(t)$.
The right half of the circle occurs between the $-\frac {\pi} 2$ and $\frac {\pi} 2$ angles. $-\pi/2 \leq t \leq \pi/2$.
2. Find dx/dy, dy/dt and f(x(t),y(t)):
$$\frac{dx}{dt} = \frac{d(4cos(t))}{dt} = -4sin(t)$$ $$\frac{dy}{dt} = \frac{d(4sin(t))}{dt} = 4cos(t)$$ $$f(x(t),y(t)) = xy^4 = (4cos(t))*(4sin(t))^4 = 1024 \space cos(t)sin^4(t)$$
3. Integrate:
$$\int_{-\pi/2}^{\pi/2} 1024 \space cos(t)sin^4(t) \sqrt {(-4sin(t))^2 + (4cos(t))^2} \space dt$$ $$\int_{-\pi/2}^{\pi/2} 1024 \space cos(t)sin^4(t) \sqrt {16(sin^2(t) + cos(t)^2)} \space dt$$ $$\int_{-\pi/2}^{\pi/2} 1024 \space cos(t)sin^4(t) (4) \space dt$$ $$ 4096 \int_{-\pi/2}^{\pi/2} \space cos(t)sin^4(t) \space dt$$
$u = sin(t)$
$u_1 = sin(-\pi/2) = -1$
$u_2 = sin(\pi/2) = 1$
$du = cos(t) dt$
$$ 4096 \int_{-\pi/2}^{\pi/2} \space u^4cos(t) \space dt$$ $$ 4096 \int_{-1}^{1} \space u^4 \space du$$ $$ 4096 [\frac 15 u^5]^{1}_{-1}$$ $$\frac{4096} 5 1^5 - (\frac {4096} 5 (-1)^5) = \frac {8192}{5} = 1638.4$$
