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, 4). Give your answer in 2 decimal places. You have to use the following formula: dz A (S) = || dz + dy 1+ dA dx

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A: We can help you with the question. I am not sure how to help you with its attachments.

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A: ∫∫∫(3x + 2y) dV=∫∫∫(3x + 2y) dxdydz = ∫z = 04∫y = -13∫x =-11(3x + 2y) dxdydz = ∫z = 04∫y = -134y…

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