Answer
$\approx 5.03 L$ of paint
Work Step by Step
Let the area of the fence be given by $A=\int h(x,y) ds$
Here, $r(t)=\lt 10 \cos t , 10 \sin t \gt$
Then, we have
$A=\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 ] \sqrt{(-10 \sin t)^2+( 10 \cos t)^2 }dt $
or, $=(4 +\cos 2t) \sqrt{100} dt$
or, $=10[4t+\dfrac{\sin 2t}{2}]_0^{2 \pi}$
Thus, $A=80 \pi m^2$
If we paint both sides of the fence, the total surface area to cover is $160 \pi m^2$.
For the given condition, we require $\dfrac{160 \pi}{100}=1.6 \pi \approx 5.03 L$ of paint
