How can we use polar coordinates to determine the volume of the solid that lies beneath the paraboloid defined by the equation z = 8 – 2x² – 2y² and above the xy-plane?

Finding the Volume Under a Paraboloid Using Polar Coordinates

To find the volume of the solid that lies beneath the paraboloid given by the equation z = 8 – 2x² – 2y² and above the xy-plane, we can simplify our calculations by switching to polar coordinates. This method helps to handle circular symmetry efficiently.

Step 1: Understanding the Surface and Volume

The paraboloid opens downward and intersects the xy-plane when z = 0:

0 = 8 – 2x² – 2y²
Simplifying this, we find:

2x² + 2y² = 8
x² + y² = 4

This equation describes a circle of radius 2 on the xy-plane. Thus, our volume calculation will be constrained within this circle.

Step 2: Converting to Polar Coordinates

In polar coordinates, we substitute:

x = r cos(θ)
y = r sin(θ)

Then, the equation for z becomes:

z = 8 – 2(r cos(θ))² – 2(r sin(θ))²

z = 8 – 2r²(cos²(θ) + sin²(θ))

Since extit{cos²(θ) + sin²(θ) = 1}, we simplify to:

z = 8 – 2r²

Step 3: Setting Up the Volume Integral

The volume V can be calculated using the double integral in polar coordinates:

V = ∫∫_D z \, r \, dr \, dθ

Where D is the circular area defined by the radius 2:

0 ≤ r ≤ 2
0 ≤ θ ≤ 2π

Substituting for z gives:

V = ∫₀²π ∫₀² (8 – 2r²) r \, dr \, dθ

Step 4: Evaluating the Integral

Now we compute the inner integral:

∫₀² (8r – 2r³) \, dr

Calculating the inner integral:

= [4r² – 0.5r⁴]₀²

= [4(2)² – 0.5(2)⁴] – 0

= [16 – 8] = 8

Now, substituting back into the outer integral:

V = ∫₀²π 8 \, dθ

This evaluates to:

= 8(2π) = 16π

Conclusion

Hence, the volume of the solid that lies underneath the paraboloid and above the xy-plane is:

V = 16π cubic units.

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