By using the double integral , find surface area of the portion of the surface z = sqrt (4 -x^2) that lies above rectangle in xy -plane whose coordinate dsatisfy 0<x<1 and 0<y<4 ...

I have sketched out the diagram , but i found that , it's merely 2 plane surfaces , it's not a 3d object formed by 2 surfaces , right ? So, we just need to find the respective surface area and add 2 things up ?

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