the integral of vector is also a vector , so
\int_{V} \nabla\times\vec{A} dV = (\int_{V} \nabla\times\vec{A} dV)_{x}\vec{i} +(\int_{V} \nabla\times\vec{A} dV)_{y}\vec{j} +(\int_{V} \nabla\times\vec{A} dV)_{z}\vec{k}
my idea is prove each component of \vec{i},\vec{j},\vec{k} in LHS is equal to that of RHS, than we can say the LHS = RHS
first, \nabla\times\vec{A} = (\frac{\partial A_{z}}{\partial y} -\frac{\partial A_{y}}{\partial z})\vec{i} - (\frac{\partial A_{z}}{\partial x} -\frac{\partial A_{x}}{\partial z})\vec{j} + (\frac{\partial A_{y}}{\partial x} -\frac{\partial A_{x}}{\partial y})\vec{kj}
so, the z component of \nabla\times\vec{A} is \frac{\partial A_{y}}{\partial x} -\frac{\partial A_{x}}{\partial y}
(\int_{V} \nabla\times\vec{A} dV)_{z} = \int_{V} \frac{\partial A_{y}}{\partial x} -\frac{\partial A_{x}}{\partial y} dV=\int_{V}\frac{\partial A_{y}}{\partial x} -\frac{\partial A_{x}}{\partial y} dxdydz
= \int_{V}\frac{\partial A_{y}}{\partial x}dxdydz - \int_{V}\frac{\partial A_{x}}{\partial y} dxdydz =\int A_{y}dydz - \int A_{x} dxdz
The z component of LHS is \int A_{y}dydz - \int A_{x} dxdz , now prove the z component of RHS is equal to above...
RHS = - \int_{S} \vec{A}\times\vec{n}dS = -\int_{S} \vec{A}\times d \vec{S}
what this integral do is sum up all little \vec{A}\times\Delta\vec{S} on a closed surface, we can compute the x,y,z direction of \vec{A}\times\Delta\vec{S} first, and put the integral signs back later..
\vec{A}\times\Delta\vec{S} is just and vector, and the cross product of this vector is simple:
\vec{A}\times\Delta\vec{S} = (\vec{A}\times\Delta\vec{S})_{x} \vec{i} +(\vec{A}\times\Delta\vec{S})_{y} \vec{j} +(\vec{A}\times\Delta\vec{S})_{z} \vec{k}
= (A_{y}\Delta S_{z} - A_{z}\Delta S_{y})\vec{i} - (A_{x}\Delta S_{z} - A_{z}\Delta S_{x})\vec{j} + (A_{x}\Delta S_{y} - A_{y}\Delta S_{x}) \vec{k}
the z component of \vec{A}\times\Delta\vec{S} is A_{x}\Delta S_{y} - A_{y} \Delta S_{x}
OK, \Delta S_{x} is the surface area that point to the direction \vec{x}, that means it is parallel to the y,z plane, therefore,
\Delta S_{x} = \Delta y \Delta z, for the same reason, \Delta S_{y} = \Delta x \Delta z, and \Delta S_{z} = \Delta x \Delta y
A_{x}\Delta S_{y} - A_{y} \Delta S_{x} become
A_{x} \Delta x \Delta z - A_{y} \Delta y \Delta z
put the integral sign back and become:
- (\int A_{x} dxdz - \int A_{y} dydz) = \int A_{y}dydz - \int A_{x}dxdz = LHS
after you done with the z component, you can argue that the x and the y component is also equal