椭圆面积公式,微积分证法

2025-03-15 05:26:55
推荐回答(2个)
回答1:

椭圆面积公式:
S=π(圆周率)×a×b(其中a,b分别是椭圆的长半轴,短半轴的长). 或S=π(圆周率)×A×B/4(其中A,B分别是椭圆的长轴,短轴的长).

首先用Newton-Leibniz公式证明了微积分第一基本定理,然后又将变上限积分函数Φ(x)=∫xaf(t)dt在[a,b]上应用Lagrange中值定理,证明了积分中值定理,亦证明了积分中值定理的中间点与微分中值定理的中间点是相一致的,从而可使微积分教学更加灵活。

别忘了给分哈

回答2:

因为两轴焦点在0点,所以椭圆的面积可以分为4个相等的部分,分别是+x+y、-x+y、-x-y、+x-y四个区域,所以只要求出一个象限间所夹的面积,然后再乘以4就可以得到整个椭圆的面积。拣最简单的来吧,先求第一象限所夹部分的面积。
根据定积分的定义及图形的性质,我们可以把这部分图形无限分为底边在x轴上的小矩形,整个图形的面积就等于这些小矩形面积和的极限。现在应用元素法,在图 形中任找取一点,然后再取距这点距离无限近的另一个点,这两点间的距离记做dx,然后取以dx为底边,两点分别对应的y为高,与曲线相交够成的封闭的小矩 形的面积s,显然,s=y*dx
现在求s的定积分,即大图形的面积S,S=∫[0:a]ydx 意思是求0 到 a上y关于x的定积分
步骤:(第一象限全取正,后面不做说明)
S=∫[0:a]ydx=∫[0:a]|sqr(b^2-b^2*x^2/a^2)|dx
设 x^2/a^2=sin^2t 则
∫[0:a]|sqr(b^2-b^2*x^2/a^2)|dx=∫[0:pi/2]b*cost d(a*sint) pi=圆周率
∫[0:pi/2]b*cost d(a*sint)=∫[0:pi/2]b*a*cos^2t dt
cos^2t=1-sin^2t
∫[0:pi/2]b*a*cos^2t dt =[a*b*t](0:pi/2)-∫[0:pi/2]b*a*sin^2t dt
这里需要用到一个公式:∫[0:pi/2]f(sinx)dx=∫[0:pi/2]f(cosx)dx
证明如下 sinx=cos(pi/2-x) 设u=pi/2-x 则
∫[0:pi/2]f(sinx)dx=∫[pi/2:0]f(cosu)d(pi/2-u)= -∫[0:pi/2]f(sinu)d(pi/2-u)=∫[0:pi/2]f(sinu)du=∫[0:pi/2]f(sinx)dx

则∫[0:pi/2]b*a*cos^2t dt =[a*b*t](0:pi/2)-∫[0:pi/2]b*a*sin^2t dt=a*b*(pi/2)-∫[0:pi/2]b*a*cos^2t dt
那么 2*∫[0:pi/2]b*a*cos^2t dt=a*b*(pi/2)
则S=a*b*(pi/4)
椭圆面积S_c=a*b*pi
可见椭圆面积与坐标无关,所以无论椭圆位于坐标系的哪个位置,其面积都等于半长轴长乘以半短轴长乘以圆周率

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