To find the area bounded by the graphs, we can first graph to find what is needed:

We see that to find the bounded area we can integrate with respect to y from y=0 to y=2. Also, due to the symmetry of the functions, we see that the area is twice the integral of positive values for x.

This means that after isolating x in the quadratic, we get:

`A=2int_0^2sqrt{y+2}dy` let `u=y+2` then `du=dy` and the limits go from 2 to 4

`=2int_2^4u^{1/2}du` now use power law for integrals

`=2 cdot 2/3 u^{3/2}|_2^4`

`=4/3(4^{3/2}-2^{3/2})`

` ` `=4/3(8-2sqrt2)`

**The area of the bounded region is `4/3(8-2sqrt2)` .**

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