Find the volume of the region bounded above by the paraboloid z equals 3 x squared plus 4 y squared and below by the square r: minus 2 less than or equals x less than or equals 2 comma negative 2 less than or equals y less than or equals 2.

149.333 cubic units.

Step-by-step explanation:

Volume of the space between a 3-D structure and a plane surface/shape (2-D) is given by the double integral of the 3-D shape evaluated definitely along the limits of the planar surface/shape.

V = ∫∫ f(x, y) dx dy with the integral evaluated along the limits of x and y for the 2-D figure.

For this problem question, we are required to evaluate the volume of the region bounded by the paraboloid z = f(x, y) = 3x² + 4y² and the square r: -2 ≤ x ≤ 2, -2 ≤ y ≤ 2

Volume = ∫²₋₂ ∫²₋₂ f(x, y) dx dy = ∫²₋₂ ∫²₋₂ [(3x² + 4y²) dx] dy

Evaluating the double integral one by one, with respect to x first

Volume = ∫²₋₂ [x³ + 4xy²]²₋₂ dy = ∫²₋₂ {[2³ + 4(2)y²] - [(-2)³ + 4(-2)y²]} dy

Volume = ∫²₋₂ [(8 + 8y²) - (-8 - 8y²)] dy

= ∫²₋₂ [16 + 16y²] dy = [16y + (16y³/3)]²₋₂

= [16(2) + 16(2³)/3] - [16(-2) + 16(-2)³/3]

= [32 + 42.667] - [-32 - 42.667]

= 74.667 + 74.667 = 149.333 cubic units.

Hope this Helps!!!

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step-by-step explanation:

what are you trying to solve here

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