Question

关于一个2次定积分问题：

Answer 1

容易看出积分区间是个矩形,相对于x轴和y轴都对称

被积函数同时是x和y的偶函数,因此根据对称性

原式=4∫[0->h]∫[0->L] (x²+y²)^(1/2) dxdy

当h=0或者L=0时,积分为0；当h,L都不为0时

可以把积分化到极坐标形式

原式=4∫[0->arctan(h/L)]∫[0->L/cosθ] r²drdθ+4∫[arctan(h/L)->π/2]∫[0->h/sinθ] r²drdθ

=4∫[0->arctan(h/L)] L³/(3cos³θ)dθ+4∫[arctan(h/L)->π/2] h³/(3sin³θ)dθ

=(4L³/3)∫[0->arctan(h/L)] 1/(1-sin²θ)²dsinθ - (4h³/3)∫[arctan(h/L)->π/2] 1/(1-cos²θ)²dcosθ

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下面来求∫1/(1-t²)²dt型的不定积分

欲求它,首先要求∫1/(1-t²)dt

∫1/(1-t²)dt=(1/2)[∫1/(1-t)dt+∫1/(1+t)dt]=(1/2)ln|(1+t)/(1-t)|+C

∫1/(1-t²)dt=t/(1-t²)-∫2t²/(1-t²)²dt=t/(1-t²)+2[∫(1-t²)/(1-t²)²dt -∫1/(1-t²)²dt]

-∫1/(1-t²)dt=t/(1-t²)-2∫1/(1-t²)²dt

∫1/(1-t²)²dt=(1/2)[t/(1-t²)+∫1/(1-t²)dt]=t/2(1-t²)+(1/4)ln|(1+t)/(1-t)|+C

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设函数f(t)=t/2(1-t²)+(1/4)ln|(1+t)/(1-t)|

原式=(4L³/3)[f(sin(arctan(h/L)) - f(sin0)] - (4h³/3)[f(cos(π/2)) - f(cos(arctan(h/L))]

易知f(0)=0,sin(arctan(h/L))=h/√(h²+L²),cos(arctan(h/L))=L/√(h²+L²)

原式=(4L³/3)f(h/√(h²+L²)) + (4h³/3)f(L/√(h²+L²))

故结果为：

1.当h=0或L=0时,积分为0；

2.当h,L都不为0时：积分=

(4/3)Lh√(h²+L²)+(1/3)L³ln[(√(h²+L²)+h)/(√(h²+L²)-h)]+(1/3)h³ln[(√(h²+L²)+L)/(√(h²+L²)-L)]

想问一句,楼主你有答案吗?