高等数学第六小题,求详细过程
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(6) 积分区域关于 y 轴对称,
x 的奇函数 e^√(x^2+y^2)sin(xy) 积分为 0
x+y= 3 化为 r = 3/(cost+sint)=(3/√2)csc(t+π/4)
I = 2∫<0,π/2>dt∫<0,√2> (2-r^2)rdr
+ 2∫<0,π/2>dt∫<√2, (3/√2)csc(t+π/4)> (r^2-2)rdr
= π[r^2-r^4/4]<0,√2>
+ 2∫<0,π/2>dt[r^4/4-r^2]<√2, (3/√2)csc(t+π/4)>
= π + 2∫<0,π/2>(9/2)[csc(t+π/4)]^2[(9/8){csc(t+π/4)]^2-1}dt
= π - 9∫<0,π/2>{(9/8)[cot(t+π/4)]^2+1/8}dcot(t+π/4)
= π - (9/8)[3{cot(t+π/4)}^3 + cot(t+π/4)]<0,π/2>
= π - (9/8)[3{cot(t+π/4)}^3 + cot(t+π/4)]<0,π/2>
= π + 9
x 的奇函数 e^√(x^2+y^2)sin(xy) 积分为 0
x+y= 3 化为 r = 3/(cost+sint)=(3/√2)csc(t+π/4)
I = 2∫<0,π/2>dt∫<0,√2> (2-r^2)rdr
+ 2∫<0,π/2>dt∫<√2, (3/√2)csc(t+π/4)> (r^2-2)rdr
= π[r^2-r^4/4]<0,√2>
+ 2∫<0,π/2>dt[r^4/4-r^2]<√2, (3/√2)csc(t+π/4)>
= π + 2∫<0,π/2>(9/2)[csc(t+π/4)]^2[(9/8){csc(t+π/4)]^2-1}dt
= π - 9∫<0,π/2>{(9/8)[cot(t+π/4)]^2+1/8}dcot(t+π/4)
= π - (9/8)[3{cot(t+π/4)}^3 + cot(t+π/4)]<0,π/2>
= π - (9/8)[3{cot(t+π/4)}^3 + cot(t+π/4)]<0,π/2>
= π + 9
追问
是2派+9
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