Answer:
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!!!
