Base of a circular fence with radius 10 m is given by x = 10 cos(t), y = 10 sin(t). the height of the fence at position (x, y) is given by the function h(x, y) = 3 + 0.03(x2 − y2), so the height varies from 0 m to 6 m. suppose that 1 l of paint covers 100 m2. determine how much paint you will need if you paint both sides of the fence. (round your answer to two decimal places.)

The fence may look like a cylinder, but it is not, since the height varies.
Therefore, you need to find how the height varies with the only parameter you know: the angle t. Then, you can integrate to find the area of the fence.

First, you need to write h(x,y) only as a function of x (or y, you can choose). Since the base is circular, you have:
x² + y² = 100
y² = 100 - x²

You can now substitute:
h(x) = 3 + 0.03(x² - y²)
       = 3 + 0.03[x² - (100 - x²)]
       = 3 + 0.03(2x² - 100)
       = 3 + 0.06(x² - 50)

Now, you have to write h as a function of t; you know:
x = 10·cos(t)
Therefore:
h(t) = 3 + 0.06[(10·cos(t)² - 50]
      = 3 + 0.06[100·cos²(t) - 50]
      = 3 + 6·cos²(t) - 3
      = 6·cos²(t)
This function is correct because we know that the height varies between 0 and 6m.

In order to find the area, you need to integrate:
[tex] \int\limits^{2\pi} _0 {6cos^{2}(t) } \, dt [/tex] =
[tex] 6(1/2)[sin(t)cos(t) + t]_{0}^{2 \pi } [/tex] =
6·(1/2)[sin(2π)cos(2π) + 2π] - 6·(1/2)[sin(0)cos(0) + 0] =
6·(1/2)(2π) =
6π 

This is one side of the fence, therefore the total area to be painted is 12π m². You need 1l for 100m², therefore:
(1 / 100) · 12π = 0.38 l

Hence, you will need 0.38 liters of paint to cover both sides of the fence.