To find the volume of the region below the graph of f(x, y) = 16 - x² - y² and above the xy-plane in the first octant, we can use a triple integral in polar coordinates. Here are the steps:

1. Convert the function to polar coordinates: f(r, θ) = 16 - r², where r² = x² + y².

2. Set up the limits of integration. Since we're only considering the first octant, 0 ≤ θ ≤ π/2, 0 ≤ r ≤ 4 (since r² = x² + y² ≤ 16), and 0 ≤ z ≤ 16 - r².

3. Set up the triple integral: ∫ from 0 to π/2 ∫ from 0 to 4 ∫ from 0 to 16 - r² r dz dr dθ. The extra factor of r is due to the Jacobian determinant when converting from Cartesian to polar coordinates.

4. Evaluate the innermost integral: ∫ from 0 to π/2 ∫ from 0 to 4 [rz] from 0 to 16 - r² dr dθ = ∫ from 0 to π/2 ∫ from 0 to 4 r(16 - r²) dr dθ.

5. Evaluate the middle integral: ∫ from 0 to π/2 [4r² - r⁴/4] from 0 to 4 dθ = ∫ from 0 to π/2 (64 - 64) dθ = 0.

6. Evaluate the outermost integral: [0] from 0 to π/2 = 0.

So, the volume of the region below the graph and above the xy-plane in the first octant is 0.

Question

To find the volume of the region below the graph of f(x, y) = 16 - x² - y² and above the xy-plane in the first octant, we can use a triple integral in polar coordinates. Here are the steps:

1. Convert the function to polar coordinates: f(r, θ) = 16 - r², where r² = x² + y².

2. Set up the limits of integration. Since we're only considering the first octant, 0 ≤ θ ≤ π/2, 0 ≤ r ≤ 4 (since r² = x² + y² ≤ 16), and 0 ≤ z ≤ 16 - r².

3. Set up the triple integral: ∫ from 0 to π/2 ∫ from 0 to 4 ∫ from 0 to 16 - r² r dz dr dθ. The extra factor of r is due to the Jacobian determinant when converting from Cartesian to polar coordinates.

4. Evaluate the innermost integral: ∫ from 0 to π/2 ∫ from 0 to 4 [rz] from 0 to 16 - r² dr dθ = ∫ from 0 to π/2 ∫ from 0 to 4 r(16 - r²) dr dθ.

5. Evaluate the middle integral: ∫ from 0 to π/2 [4r² - r⁴/4] from 0 to 4 dθ = ∫ from 0 to π/2 (64 - 64) dθ = 0.

6. Evaluate the outermost integral: [0] from 0 to π/2 = 0.

So, the volume of the region below the graph and above the xy-plane in the first octant is 0.

Knowee AI · Accepted Answer

To find the volume of the region below the graph of f(x, y) = 16 - x² - y² and above the xy-plane in the first octant, we can use a triple integral in polar coordinates. Here are the steps:

1. Convert the function to polar coordinates: f(r, θ) = 16 - r², where r² = x² + y².

2. Set up the limits of integration. Since we're only considering the first octant, 0 ≤ θ ≤ π/2, 0 ≤ r ≤ 4 (since r² = x² + y² ≤ 16), and 0 ≤ z ≤ 16 - r².

3. Set up the triple integral: ∫ from 0 to π/2 ∫ from 0 to 4 ∫ from 0 to 16 - r² r dz dr dθ. The extra factor of r is due to the Jacobian determinant when converting from Cartesian to polar coordinates.

4. Evaluate the innermost integral: ∫ from 0 to π/2 ∫ from 0 to 4 [rz] from 0 to 16 - r² dr dθ = ∫ from 0 to π/2 ∫ from 0 to 4 r(16 - r²) dr dθ.

5. Evaluate the middle integral: ∫ from 0 to π/2 [4r² - r⁴/4] from 0 to 4 dθ = ∫ from 0 to π/2 (64 - 64) dθ = 0.

6. Evaluate the outermost integral: [0] from 0 to π/2 = 0.

So, the volume of the region below the graph and above the xy-plane in the first octant is 0.

Find the volume of the region below the graph of f (x, y) = 16 − x2 − y2 and above thexy-plane in the ﬁrst octant. Hint: convert to polar coordinates

Question

Find the volume of the region below the graph of $f (x, y) = 16 − x^2 − y^2$ and above the xy-plane in the first octant.

Solution

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