Final answer:
The volume of the region between the graph of f(x,y) = 36 - x² - y² and the xy-plane is 432π cubic units.
Step-by-step explanation:
To find the volume of the region between the graph of the function f(x,y) = 36 - x² - y² and the xy-plane, we can use a double integral. The volume V is given by the double integral of the function f(x,y) over the region R in the xy-plane.
The region R is defined by the domain of the function f(x,y), which is a circle centered at the origin with radius 6. Therefore, we can express R as 0 ≤ x ≤ 6 and -√(36-x²) ≤ y ≤ √(36-x²).
The volume V is then given by:
V = ∬R f(x,y) dA
Where dA represents an infinitesimal area element in the xy-plane.
Substituting f(x,y) = 36 - x² - y² into the double integral, we get:
V = ∬R (36 - x² - y²) dA
To evaluate this double integral, we can use polar coordinates. In polar coordinates, the region R is described by 0 ≤ r ≤ 6 and 0 ≤ θ ≤ 2π.
The transformation to polar coordinates involves substituting x = rcos(θ) and y = rsin(θ), and replacing dA with r dr dθ.
The volume V in polar coordinates becomes:
V = ∫02π ∫06 (36 - r²) r dr dθ
Evaluating this double integral gives us the volume of the region between the graph of f(x,y) and the xy-plane.
Final Answer:
V = 432π cubic units