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Multivariable & Vectorhard
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Calculate the surface area of the part of the surface z=23(x3/2+y3/2)z = \frac{2}{3}(x^{3/2} + y^{3/2})z=32​(x3/2+y3/2) that lies above the square region D={(x,y)∣0≤x≤1,0≤y≤1}D = \{(x, y) \mid 0 \le x \le 1, 0 \le y \le 1\}D={(x,y)∣0≤x≤1,0≤y≤1}.