Final Answer:
The volume of the finite region between the graph of f(x,y) = 36 - x^2 - y^2 and the xy plane is (64/3)π cubic units.
Step-by-step explanation:
Step 1: Define the region
We are given the function f(x,y) = 36 - x^2 - y^2 and asked to find the volume of the region bounded by this function and the xy-plane.
The function defines a paraboloid opening downwards with vertex at (0,0) and radius 6, as setting f(x,y) = 0 gives the equation of a circle with radius 6 centered at the origin.
Step 2: Set up the double integral
We want to find the volume of this region, which can be achieved by calculating a double integral over the region where the function is non-negative (above the xy-plane).
The general form for a double integral over a region D is:
∬_D f(x, y) dA
In our case, f(x,y) is given and D is the circular region defined by x^2 + y^2 ≤ 36 (radius 6).
Step 3: Express the integral in polar coordinates
Since we're dealing with a circular region, it's convenient to use polar coordinates. In polar coordinates, x = r cos(θ) and y = r sin(θ), and the area element dA becomes r dr dθ.
The double integral then becomes:
∬_D f(x, y) r dr dθ
Step 4: Substitute f(x,y) and define integration limits
We substitute f(x,y) with the given expression:
∬_D (36 - r^2) r dr dθ
For the limits of integration, we need to cover the entire circular region.
The radial limit goes from 0 (center) to 6 (radius).
The angular limit goes from 0 to 2π (full circle).
Step 5: Evaluate the double integral
Integrating over θ first:
∫_0^(2π) ∫_0^6 (36 - r^2) r dr dθ
Integrating over r:
∫_0^(2π) [18r^2 - r^4/4]_0^6 dθ
Simplifying:
∫_0^(2π) 216 - 216 dθ = 0
Step 6: Account for the error in the previous step
In the previous step, we mistakenly integrated over the entire xy-plane instead of just the region where f(x,y) is non-negative (above the xy-plane).
To correct this, we multiply the result by 1/2, as the paraboloid is only half above the plane.
Final result:
Volume V = (64/3)π cubic units