Find the surface area of the part of the surface z = x2 + 3y that lies above the triangular region T in the xy- plane with the vertices (0, 0), (1, 0), and (1, 6). Give your answer in 2 decimal...

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Extracted text: Find the surface area of the part of the surface z = x2 + 3y that lies above the triangular region T in the xy- plane with the vertices (0, 0), (1, 0), and (1, 6). Give your answer in 2 decimal places. You have to use the following formula: %3D əz 2 dA ду az 2 A(S) = [[ / 1 + ax