?? 2? 0d?? 3 0?12?r?z?2?2? r3 4?r2dr? 31?2 2? 0d?? 3 05?r?3?4r?r??dr.
9???214r6?13??2??2r?r???. ?2454? 04?146.解：因为z?222a?x?y,故z???xxa?x?yy222222,z???yaya?x?y222, 1?z??z??2x2y1?2x2222a?x?y22?a?x?y2?a?x?y222.
221于是 ??d??zS??Dxya?x?y?aa?x?y222dxdy???adxdy?a??a?hDxy?.
47.解：S是x2?y2?z2?a2(x?0,y?0)分解为两部分:
S1:x?y?z?aS2:x?y?z222222?x?0,y?0,z?0?,
2?a?x?0,y?0,z?0?.
2故 ??zdxdy?S??zdxdy???zdxdy
S1S2???Dxya?x?ydxdy?222????Dxya?x?y a222?dxdy
2?2??Dxya?x?ydxdy?222? ?2 0d??ra?rdr? 0216a?.
3?2?1??2???y2?2y?y2??y2?22???3y???f?dxdydz 48.解：原式= ????3x??f?????2??3z????zzzyzz??????V?????? ?3???(x?y?z)dxdydz ?3?V 222 2? 0d?? ?4 0d??rrsin?dr
a b2244?2?b?a ?6??sin?d??rdr?6??1?. ??? 0 a25?? z ?4 b4
o x
49.解:(Ⅰ).画出积分区域
z
21
y
o y
x
(Ⅱ).原式=????x?y?z?dxdydz?222V? 2? 02d??d??r.rsin?dr? 0 02 ? a224?5a.
550.解：由Gauss公式，得I???xyS2dydz?yzdzdx?zxdxdy????(xV222?y?z)dV，由广
?x?arsin?cos??义球坐标变换 ?y?brsin?sin?, J(r,?,?)?abcr?z?crcos??2sin?，得
I??2?0d??d??(arsin?cos??brsin?sin??crcos?)abcr00?1222222222222sin?dr
?abc54?15
??2?0d?2??0(asin?cos??bsin?sin??ccos?)sin?d?22222222
22abc(a?b?c).
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