## Calculus: Early Transcendentals 8th Edition

$\approx 5.03 L$ of paint
Surface Area, $A=\int h(x,y) ds=\int_0^{2 \pi} h(r(t))|r'(t)| dt=\int_0^{2 \pi} [4 +0.01 ((10 \cos u)^2)-(10 \sin u )^2 ] \sqrt{(-10 \sin t)^2+( 10 \cos t)^2 }dt$ Thus, we have $A=(4 +\cos 2t) \sqrt{100} dt=10[4t+\dfrac{\sin 2t}{2}]_0^{2 \pi}=80 \pi m^2$ When we want to paint both sides of the fence, then the total surface area to cover becomes $160 \pi m^2$. and we need $\dfrac{160 \pi}{100}=1.6 \pi \approx 5.03 L$ of paint