问题描述：
高数微积分题一道
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 varies from 3 m to 5 m such that,at position (x,y),the height is given by
the function h(x,y) = 4 + 0.01(x2 -y2).Suppose that 1 litre of paint covers 100 m2.Sketch
the fence and determine how much paint you will need if you paint both sides of the fence.
Ans:5 litre
请具体讲一下思路,万分感激!
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 varies from 3 m to 5 m such that,at position (x,y),the height is given by
the function h(x,y) = 4 + 0.01(x2 -y2).Suppose that 1 litre of paint covers 100 m2.Sketch
the fence and determine how much paint you will need if you paint both sides of the fence.
Ans:5 litre
请具体讲一下思路,万分感激!
问题解答：
我来补答
如图进行微分.微元面积等于高度h乘以微元弧长.
dS=h*dl=h*R*dθ
所以
S=∫Rhdθ（积分范围从0到2π）
=∫10*（4 + ((cosθ)^2 -(sinθ)^2）)dθ
剩下的你会算了吧
dS=h*dl=h*R*dθ
所以
S=∫Rhdθ（积分范围从0到2π）
=∫10*（4 + ((cosθ)^2 -(sinθ)^2）)dθ
剩下的你会算了吧
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