Question

What's the value of the integral? # intintint_omega(sqrt(1-x^2-4y^2-9z^3)dxdydz) Omega:{ x^2+4y^2+9z^3<=1; x,y,z>=0}#

Answer 1

No idea.

Explanation:

Let

`I=intintintsqrt(1-x^2-4y^2-9z^3)dxdydz`

Sine `x^2+4y^2+9z^3≤1`, take the series expansion of the square root:

`I=intintintsum_(n=0)^oo((1/2),(n))(-x^2-4y^2-9z^3)^ndxdydz`

Apply the trinormial expansion:

`I=intintintsum_(n=0)^oo((1/2),(n))(-1)^nsum_(i+j+k=n)(n!)/(i!*j!*k!)(x^2)^i(4y^2)^j(9z^3)^kdxdydz`

Separate the variables:

`I=sum_(n=0)^oosum_(i+j+k=n)((1/2),(n))(n!)/(i!*j!*k!)(-1)^n4^j9^k(intx^(2i)dx)(inty^(2j)dy)(intz^(3k)dz)`

Integrate directly:

`I=sum_(n=0)^oosum_(i+j+k=n)((1/2),(n))(n!)/(i!*j!*k!)(-1)^n4^j9^k(x^(2i+1)/(2i+1))(y^(2j+1)/(2j+1))(z^(3k+1)/(3k+1))+C`

Hence

`I=sum_(n=0)^oosum_(i+j+k=n)((1/2),(n))(n!)/(i!*j!*k!)((-1)^n4^j9^k)/((2i+1)(2j+1)(3k+1))x^(2i+1)y^(2j+1)z^(3k+1)+C`