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

User Divy Soni
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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

User Byron Gavras
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3 votes

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
\(\theta\):


\[x = r \cos \theta\ \\\\[y = r \sin \theta\]

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.

User Jtjacques
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8.5k points

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