Question

Find the surface area of the region S on the plane z=3x+7y such that 0≤x≤30 and 0≤y≤5 by finding a parameterization of the surface and then calculating the surface area.

Answer 1

Answer:

$150\sqrt{59}$

Step-by-step explanation:

$z = 3x + 7y\\\\\frac{dz}{dx} = 3\:\:\: and \:\:\: \frac{dz}{dy} =7$

$Area = \int\limits^5_0 \int\limits^{30}_0 \sqrt{1 + (\frac{dz}{dx})^2+ (\frac{dz}{dy})^2}\:dx\:dy\\\\Area = \int\limits^5_0 \int\limits^{30}_0 \sqrt{1 + 9+ 49}\:dx\:dy=\sqrt{59}*30*5=150\sqrt{59}$