   Chapter 16.2, Problem 48E

Chapter
Section
Textbook Problem

The 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) = 4 + 0.01(x2 − y2), so the height varies from 3 m to 5 m. Suppose that 1 L of paint covers 100 m2. Sketch the fence and determine how much paint you will need if you paint both sides of the fence.

To determine

To sketch: The circular fence and paint required to pain both sides of the fence.

Explanation

Given data:

The radius of the circular fence is 10 m.

The parameters are given as follows.

x=10cost,y=10sint

The height of the circular fence is varied from 3 m to 5 m and the function of height is given as follows.

h(x,y)=4+0.01(x2y2) (1)

1 L of paint covers the area of 100m2 of the fence.

Formula used:

Write the expression to find area of the fence.

A=Ch(x,y)ds (2)

Here,

C is the base of the fence.

Write the expression to find ds .

ds=(dxdt)2+(dydt)2dt (3)

Write the expression to find paint required.

Paint required=(Paint required for 1 m2)(Total Area of the fence) (4)

Consider the base of the circular fence lies in the xy-plane.

Consider the center of the circular fence lies in the origin with the height z=h(x,y) .

From equation (1) and the given data (height varies from 3 m to 5 m), notice that the fence has highest height of 5m when the parameter y=0 and the fence has lowest height of 3 m when the parameter x=0 .

From equation (1), the height of the fence is 4 m when the parameter y=±x .

Consider the parameter t as u. and the write the parameters x and y as follows.

x=10cosu,y=10sinu,0u2π

As the z=h(x,y) , writhe the parameter z as follows.

z=v[4+0.01(x2y2)],0v1

Substitute 10cosu for x and 10sinu for y .

z=v{4+0.01[(10cosu)2(10sinu)2]},0v1=v[4+0.01(100)(cos2usin2u)],0v1=v(4+cos2u),0v1 (cos2θsin2θ=cos2θ)

From the parameters, write the parametric equations of the fence as follows.

x=10cosu,y=10sinu,z=v(4+cos2u),0u2π,0v1

Use the parametric equations and sketch the fence as shown in Figure 1.

Thus, the fence is sketched for the given data.

Calculation of ds :

Substitute 10cost for x and 10sint for y in equation (3),

ds=[ddt(10cost)]2+[ddt(10sin)t]2dt=(10sint)2+(10cost)2dt=100(sin2t+cos2t)dt=10(1)dt {sin2θ+cos2θ=1}

ds=10dt

Calculation of area of fence:

Substitute [4+0

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