Which of the following sets contains an element that cannot be classified as rational?
1 answer:
You might be interested in
By the divergence theorem, the surface integral given by
(where the integral is computed over the entire boundary of the surface) is equivalent to the triple integral
where
is the volume of the region 
bounded by 
.
You have
![\implies \nabla\cdot\mathbf F=\dfrac{\partial}{\partial x}[x^2y]+\dfrac{\partial}{\partial y}[xy^2]+\dfrac{\partial}{\partial z}[4xyz]=8xy](https://tex.z-dn.net/?f=%5Cimplies%20%5Cnabla%5Ccdot%5Cmathbf%20F%3D%5Cdfrac%7B%5Cpartial%7D%7B%5Cpartial%20x%7D%5Bx%5E2y%5D%2B%5Cdfrac%7B%5Cpartial%7D%7B%5Cpartial%20y%7D%5Bxy%5E2%5D%2B%5Cdfrac%7B%5Cpartial%7D%7B%5Cpartial%20z%7D%5B4xyz%5D%3D8xy)
and so the integral reduces to
(where the integral is computed over the entire boundary of the surface) is equivalent to the triple integral
where
You have
and so the integral reduces to