Answer
$\approx 5.03 L$ of paint
Work Step by Step
$S=\int h(x,y) ds=\int_0^{2 \pi} h(r(t))|r'(t)| dt$
$=\int_0^{2 \pi} [4 +0.01 ((10 \cos u)^2)-(10 \sin u )^2 ] \times [ \sqrt{(-10 \sin t)^2+( 10 \cos t)^2 }dt] $
$=(4 +\cos 2t) \times \sqrt{100} dt$
$=10 \times [4t+\dfrac{\sin 2t}{2}]_0^{2 \pi}$
So, $S=80 \pi m^2$
When we paint both sides of the fence , then the total surface area has to be covered: $160 \pi m^2$.
Now, we require $\dfrac{160 \pi}{100}=1.6 \pi \approx 5.03 L$ of paint
