二重积分问题，积分换次序 二重积分 交换次序
\u4e8c\u91cd\u79ef\u5206\u95ee\u9898\u3002\u600e\u4e48\u6539\u53d8\u79ef\u5206\u6b21\u5e8f\uff1f\uff1f\u8d8a\u8be6\u7ec6\u8d8a\u597d\u3002\u753b\u51fa\u79ef\u5206\u533a\u57df\uff0c\u4f5c\u4e00\u6761\u5e73\u884c\u4e8ex\u8f74\u7684\u76f4\u7ebf\u7a7f\u8fc7\u79ef\u5206\u533a\u57df\uff0c\u4e0e\u79ef\u5206\u533a\u57df\u4ea4\u4e8e\u4e24\u70b9\uff0c\u628a\u8fd9\u4e24\u70b9\u7684x\u8868\u793a\u51fa\u6765\uff0c\u5c31\u662f\u79ef\u5206\u4e0a\u4e0b\u9650\u3002
\u5982\u56fe\u6240\u793a\uff1a
\u4e3b\u8981\u5c31\u662f\u8981\u7528y\u6765\u8868\u793ax\uff0c\u7136\u540e\u5c31\u4f1a\u6d89\u53ca\u5230\u5f00\u6839\u53f7\u7684\u6b63\u8d1f\u95ee\u9898\u662f\u5427\uff0c\u7136\u540e\u4f1a\u53d1\u73b0\uff0cx=a\u662f\u4e00\u4e2a\u5206\u754c\u7ebf\uff0c\u5de6\u8fb9\u53d6\u8d1f\uff0c\u53f3\u8fb9\u53d6\u6b63\uff0c(\u53ef\u4ee5\u5047\u8bbe\u503c\u53bb\u8bd5)\uff0c\u8fd9\u9053\u9898\u53d6\u7684\u5de6\u8fb9\uff0c\u6240\u4ee5\u4e3a\u8d1f\u3002
\u6269\u5c55\u8d44\u6599\uff1a\u4ea4\u6362\u79ef\u5206\u6b21\u5e8f\u4e3b\u8981\u662f\u6839\u636e\u539f\u6765\u7684\u79ef\u5206\u6b21\u5e8f\u753b\u79ef\u5206\u533a\u57df\u548c\u786e\u5b9a\u4e0a\u4e0b\u9650\u3002
\u987a\u53e3\u6e9c\uff1a
1\u3001\u540e\u79ef\u5148\u5b9a\u9650\uff0c\u9650\u5185\u7a7f\u6761\u7ebf\uff0c\u5148\u4ea4\u4e0b\u9650\u5199\uff0c\u540e\u4ea4\u4e0a\u9650\u89c1\uff1b
2\u3001\u5148\u79ef x\uff0c\u753b\u6a2a\u7ebf\uff08\u5e73\u884c\u4e8e x \u8f74\uff09\uff0c\u53f3\u51cf\u5de6\uff1b
3\u3001\u5148\u79ef y\uff0c\u753b\u7ad6\u7ebf\uff08\u5e73\u884c\u4e8e y \u8f74\uff09\uff0c\u4e0a\u51cf\u4e0b\u3002
\u222b\u222b_D \u221a(y - x²) dxdy
= \u222b(-1-->1) dx \u222b(0-->2) \u221a(y - x²) dy
= \u222b(-1-->1) dx \u222b(0-->2) \u221a(y - x²) d(y - x²)
= \u222b(-1-->1) (2/3)(y - x²)^(3/2) |(0-->2) dx
= \u222b(-1-->1) (2/3)(2 - x²)^(3/2) dx
= (4/3)\u222b(0-->1) (2 - x²)^(3/2) dx
\u4ee4x = \u221a2sin\u03b8\uff0cdx = \u221a2cos\u03b8d\u03b8
\u5f53x = 0\uff0c\u03b8 = 0\uff0c\u5f53x = 1\uff0c\u03b8 = \u03c0/4
= (4/3)\u222b(0-->\u03c0/4) (2 - 2sin²\u03b8)^(3/2) \u221a2cos\u03b8d\u03b8
= (4/3)(2\u221a2)(\u221a2)\u222b(0-->\u03c0/4) cos⁴\u03b8 d\u03b8
= (16/3)\u222b(0-->\u03c0/4) [(1 + cos2\u03b8)/2]² d\u03b8
= (4/3)\u222b(0-->\u03c0/4) (1 + 2cos2\u03b8 + cos²2\u03b8) d\u03b8
= (4/3)\u222b(0-->\u03c0/4) (1 + 2cos2\u03b8) d\u03b8 + (2/3)\u222b(0-->\u03c0/4) (1 + cos4\u03b8) d\u03b8
= (4/3)(\u03b8 + sin2\u03b8) |(0-->\u03c0/4) + (2/3)(\u03b8 + 1/4 \u00b7 sin4\u03b8)|(0-->\u03c0/4)
= (4/3)(\u03c0/4 + 1) + (2/3)(\u03c0/4)
= \u03c0/2 + 4/3
画图的话就是把上下限当做不等式的两边，对谁积分就是谁的范围，这一题就是：y<x<π/6; 0<y<π/6;，然后把这两个不等式画在直角坐标系上再重新调整一下，把x的范围调整成常数，把y的范围用x表示即使交换了积分次序了之后的范围了。
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