Question

Find the volume of the region between the graph of f(x,y) = 36 - x² - y² and the xy-plane?

Answer 1

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

Answer 2

The volume of the region between the graph of f(x,y) = 36 - x² - y² and the xy-plane is
`\(648\pi\)` cubic units.

To find the volume of the region between the graph of
`\(f(x, y) = 36 - x^2 - y^2\)` and the xy-plane, you're essentially looking for the volume under the surface
`\(f(x, y)\)` over some region in the xy-plane. In this case, the region is the circle
`\(x^2 + y^2 \leq 36\)` in the xy-plane, which corresponds to the domain of
`\(f(x, y)\)`.

This volume can be calculated using a double integral over the given domain.

The function
`\(f(x, y) = 36 - x^2 - y^2\)` represents a paraboloid centered at the origin with a vertex at
`\(z = 36\)` and opening downward.

The integral for the volume
`\(V\)` can be set up as follows:

`\[V = \iint\limits_D f(x, y) \, dA\]`

Where
`\(D\)` represents the region in the xy-plane defined by
`\(x^2 + y^2 \leq 36\)`, which is a circle of radius 6 centered at the origin.

The integral in polar coordinates would be suitable for this circular domain, so we can express
`\(x\) and \(y\)` in terms of
`\(r\)` and
`\(\theta\)`:

The limits of integration for
`\(r\)` would be from 0 to 6, and for
`\(\theta\)` would be from 0 to
`\(2\pi\)` (to cover the entire circle).

Now, the integral becomes:

`\[V = \iint\limits_D (36 - x^2 - y^2) \, dA\\\\ \[V = \int_0^(2\pi) \int_0^6 (36 - r^2) \cdot r \, dr \, d\theta\]`

Let's proceed with the calculation:

`\[V = \int_0^(2\pi) \left[(36r^2)/(2) - (r^4)/(4)\right]_0^6 \, d\theta\]\ \\\ V = \int_0^(2\pi) \left[18r^2 - (r^4)/(4)\right]_0^6 \, d\theta\]\ \\\ V = \int_0^(2\pi) (648 - 324) \, d\theta\]\ \\\ V = \int_0^(2\pi) 324 \, d\theta\]\ \\\ V = 324 \cdot \theta \Bigg|_0^(2\pi)\]\ \\\ V = 324 \cdot (2\pi - 0)\]\ \\\ V = 648\pi\]`

Therefore, the volume of the region between the graph of
`\(f(x, y) = 36 - x^2 - y^2\)` and the xy-plane over the circular region
`\(x^2 + y^2 \leq 36\)` is
`\(648\pi\)` cubic units.