Question

高等数学(综合题)： 三、求下列重积分(求第3、4小题)。

Answer 1

3. 联立解 y^2 = x, y = x 得交点O(0, 0), A(1, 1),
得 I = ∫∫<D>(siny/y)dxdy = ∫<0, 1>(siny/y)dy∫<y^2,y>dx
= ∫<0, 1>(siny/y)(y-y^2)dy
= ∫<0, 1>sinydy - ∫<0, 1>ysinydy
= [-cosy]<0, 1> + ∫<0, 1>ydcosy
= 1-cos1 + [ycosy]<0, 1> -∫<0, 1>cosydy
= 1-cos1 + cos1 - [siny]<0, 1> = 1 - sin1
4. 由对称性得
I = 8∫∫∫<D1>(x+y+z)dv
= 8∫<0, a>dz∫<<0, π/2>dt∫<0, a>(rcost+rsint+z)rdr
= 8∫<0, a>dz∫<<0, π/2>dt [(r^3/3)(cost+sint)+zr^2/2]<0, a>
= 8∫<0, a>dz∫<<0, π/2> [(a^3/3)(cost+sint)+za^2/2]dt
= 8∫<0, a>dz [(a^3/3)(sint-cost)+(1/2)zta^2]<<0, π/2>
= 8∫<0, a> [(2a^3/3)+(π/4)za^2]dz
= 8 [(2a^3/3)z + (π/8)a^2z^2]<0, a>
= 8[2/3+π/8]a^4 = (π+16/3)a^4

追问

第3题正确。第四题可以再进行复检一下妈?

追答

4. 由对称性得
I = 8∫∫∫(x+y+z)dv
= 8∫dz∫dt∫(rcost+rsint+z)rdr
= 8∫dz∫dt [(r^3/3)(cost+sint)+zr^2/2]
= 8∫dz∫[(1/3)(a-z)^3(cost+sint)+(1/2)z(a-z)^2]dt
= 8∫dz[(1/3)(a-z)^3(sint-cost)+(1/2)z(a-z)^2t]
= 8∫ [(2/3)(a-z)^3+(π/4)z(a-z)^2]dz
= -(16/3)∫ (a-z)^3d(a-z)
+ 2π∫(z^3-4az^2+4a^2z)dz
= -(4/3)[(a-z)^4]
+ 2π[z^4/4-(4/3)az^3+2a^2z^2]
= (4/3+11π/6)a^4

Answer 2

这个还真不知道，你可以百度搜索一下