设f(x,y)是连续函数,而D:x²+y²≤1且y>0,则∫∫f(√x²+y²)dxdy=? 函数y=f(x)是由方程x|x|+y|y|=1确定的函数,则...

\u8bbe\u51fd\u6570y=y(x)\u7531\u65b9\u7a0by=f(x^2+y^2)+f(x+y)\u786e\u5b9a,\u4e14y(0)=2,f(x)\u662f\u53ef\u5bfc\u51fd\u6570,f'(2)=1/2,f'(4)=1,\u5219f'(0)\u7684\u503c

y=f(x²+y²)+f(x+y)
y'=f'(x²+y²)\u00d7(x²+y²)'+f'(x+y)\u00d7(x+y)'
=(2x+2yy')f'(x²+y²)+(1+y')f'(x+y)
\u5f53x=0\u65f6\uff0cy=2\uff0c\u90a3\u4e48y'=(0+4y')f'(4)+(1+y')f'(2)

\u800cf'(4)=1\uff0cf'(2)=1/2\uff0c\u6240\u4ee5y'=4y'\u00d71+(1+y')\u00d7(1/2)
\u5373\uff1ay'=4y'+1/2+y'/2\uff0c\u6240\u4ee5y'=-1/7\uff0c\u5373f'(0)=-1/7

(1) x \u2265 0, y \u2265 0
x² + y² = 1
\u6b64\u4e3a\u5706\u5fc3\u5728\u539f\u70b9\uff0c\u534a\u5f84\u4e3a1\u7684\u5706\u5728\u7b2c1\u8c61\u9650\u7684\u90e8\u5206, \u51cf\u51fd\u6570

(2) x \u2265 0, y < 0
x² - y² = 1
\u6b64\u4e3aa = b = 1\uff0c\u5b9e\u8f74\u4e3ax\u8f74\u7684\u53cc\u66f2\u7ebf\u5728\u7b2c4\u8c61\u9650\u7684\u90e8\u5206, \u51cf\u51fd\u6570

(3) x < 0, y \u2265 0
y² - x² = 1
\u6b64\u4e3aa = b = 1\uff0c\u5b9e\u8f74\u4e3ay\u8f74\u7684\u53cc\u66f2\u7ebf\u5728\u7b2c2\u8c61\u9650\u7684\u90e8\u5206, \u51cf\u51fd\u6570

(3) x < 0, y < 0
\u4e0d\u5b58\u5728

用极坐标:x = rcos t; y = r sin t
∫∫f(√x²+y²)dxdy
= ∫[0,pi] dt ∫[0,1] f(r) rdr (注意:是上半圆)

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