Set up a double integral that gives the area of the surface of the graph region R. ƒ (x, y) = xª – 5xy − 9y R={(x,y):-5≤ x ≤5,- 4y <4} L₁1²₁1+(4x²-53) - (45) ° - 5x)ª dy dx Lºslª √¹ + (4x³ - 5y) - (45yª – 5x)” dy dx O lªl², v1 + (4x³− 5y)² = (45yª – 5x)ª dy dx L²lª² √¹ - (4x³-5y)² + (45y¹ − 5x)² dx dy O [²3 [², √ ¹ + (4x³ + 5y)² - (45yª + 5x)²³ dy dx

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