Respuesta :
The question is missing important equations. The complete question is:
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(x^2 − y^2), so the height varies from 0 m to 6 m. suppose that 1 L of paint covers 100 m^2. determine how much paint you will need if you paint both sides of the fence. (round your answer to two decimal places.)
Answer:
0.38 liters of paint is needed to cover both sides of the fence.
Step-by-step explanation:
Given:
Radius of circular base is, [tex]r=10\ m[/tex]
The height of the fence varies as:
[tex]h(x, y) = 3 + 0.03(x^2-y^2)[/tex]------------1
Where, 'x' and 'y' varies with 't' as:
[tex]x = 10 \cos(t), y = 10\sin (t)[/tex]
First, we need to write [tex]h(x,y)[/tex] only as a function of [tex]x\ or\ y[/tex], As the base of the fence is circular, we use the circle's equation:
[tex]x^2 + y^2 = r^2\\x^2+y^2=100[/tex]
[tex]y^2 = 100 - x^2[/tex]
Now, plugging the value of [tex]y^2[/tex] in equation (1), we get:
[tex]h(x) = 3 + 0.03(x^2 - y^2)[/tex]
[tex]h(x) =3 + 0.03[x^2 - (100 - x^2)]=3 + 0.03(2x^2 - 100)=3 + 0.06(x^2 - 50)[/tex]
Now, we know that:
[tex]x = 10\cos(t)[/tex]
Therefore, the height can be expressed in terms of 't' as:
[tex]h(t) = 3 + 0.06[(10cos(t))^2 - 50]\\h(t) = 3 + 0.06[100cos^2(t) - 50]\\h(t)= 3 + 6cos^2(t)-3\\h(t)=6cos^2(t)-------2[/tex]
In order to find the area, we need to integrate equation (2).
[tex]\int\limits^{2\pi}_0 {h(t)} \, dt\\ =\int\limits^{2\pi}_0 {6\cos^2(t)} \, dt\\\\= 6\frac{1}{2}[\sin(t)\cos(t) + t]_{0}^{2 \pi } \\=6(\frac{1}{2})[sin(2\pi)cos(2\pi) + 2\pi] - 6(\frac{1}{2})[sin(0)cos(0) + 0] \\=3(2\pi)-3(0)\\=6\pi[/tex]
The above area is the area of one side of the fence.
Therefore, the total area to be painted is 12π m².
Now, For 100 m² of area, volume of paint needed is 1 L.
So, for 1 m² of area, volume of paint needed is [tex]\frac{1}{100}\ L[/tex]
Therefore, for 12π m² of area, volume of paint needed is:
[tex]V=\frac{1}{100}\times 12\pi=0.12\pi=0.38\ L[/tex]
Hence, 0.38 liters of paint is needed to cover both sides of the fence.
