Some initial deductions around the edges already give some regions:
We can break in some more with internal clues:
A 1 that's next to two cells of the same region must be part of that region. Similarly, a 3 next to two cells of the same region must not be part of that region.
Using that logic and avoiding making groups of size more than 10 gives:
Some more clever deductions can lead to the next step:
I've marked off two areas in blue and yellow (and also drawn a temporary diagonal border line). The yellow area has 22 cells; combined with the region of 8 next to it, that makes 30, so the big 8 region must stop there.
Similarly, in the blue area, there are only 13 or 14 cells (depending on whether the 2 is up or down). Either way, at least 6 cells need to stick out, going down the left column. These six cannot be part of the nearby 0's region.
Those six cells are forced to continue this far (because if they used the 3s near the left side, they would stop short). So this drags that region down to the 1, and that 1 completes the region.
And now we can continue with deductions, since both colored regions have been satisfied.
That gives a break-in to the rest of the puzzle:
Continuing with deductions from that point, the top and left sides can be mostly resolved...
Some more easy deductions finish off the top half:
Avoiding the 3s near the bottom cutting off a region too soon gives further progress:
And finally, the unassigned 3 going left would block off the bottom region.
So the solution to the puzzle is: