设f（x，y）为连续函数，且f（x，y）=x?x2+y2≤a2f（x，y）dxdy+y2，则f（x，y）=

\u9ad8\u7b49\u6570\u5b66\u4e0b\u518c\uff0c\u8bbef(x,y)\u4e3a\u8fde\u7eed\u51fd\u6570,\u4e14 f(x,y)=x^2*e^(-y^2)-\u222b\u222bf(u,v)dudv

\u628a\u90a3\u4e2a\u79ef\u5206\u8bbe\u4e3a\u5e38\u6570\u4ee3\u5165\u8ba1\u7b97\u5373\u53ef\u3002

\u79ef\u5206\u533a\u57df\uff1a

\u2235f\uff08x\uff0cy\uff09\u662f\u95ed\u533a\u57dfx2+y2\u2264a2\u4e0a\u7684\u8fde\u7eed\u51fd\u6570\u2234\u7531\u4e8c\u91cd\u79ef\u5206\u4e2d\u503c\u5b9a\u7406\uff0c\u5f97\u222b\u222bDf(x\uff0cy)dxdy=\u03c0a2f\uff08\u03be\uff0c\u03b7\uff09\uff0c\u5176\u4e2d\uff08\u03be\uff0c\u03b7\uff09\u662f\u79ef\u5206\u533a\u57dfD\u4e2d\u7684\u67d0\u4e00\u70b9\u2234lima\u219201\u03c0a2?Df(x\uff0cy)dxdy=lima\u21920\u03c0a2\u03c0a2f(\u03be\uff0c\u03b7)\uff1dlimx\u21920f(\u03be\uff0c\u03b7)\uff1df(0\uff0c0)

因为f（x，y）连续，从而f（x，y）在积分区域x²+y²≤a²上可积，故可设

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x2+y2≤a2

f(x，y)dxdy＝A，
于是f（x，y）=xA+y²，
从而两边在区域x²+y²≤a²上积分可得如下等式：
A=

?

x2+y2≤a2

(Ax+y2)dxdy
=A

?

x2+y2≤a2

xdxdy+

?

x2+y2≤a2

y2dxdy．
利用积分区域的对称性可得，

?

x2+y2≤a2

xdxdy=0．
利用极坐标系计算可得，

?

x2+y2≤a2

y2dxdy=

∫

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