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.
