Exploring the Volume of a Solid under a Paraboloid and within a Cylinder

Mathematical analysis plays a crucial role in understanding complex shapes and volumes in three-dimensional space. In this article, we will delve into the fascinating process of finding the volume of the solid that lies under the paraboloid defined by z x^2 - y^2 and above the xy-plane while being limited by a cylinder. This exploration will involve several steps, including the essential transformations and integrations necessary to arrive at the final answer. Let's embark on this journey through geometry and calculus!

Step 1: Understanding the Cylinder

The equation of the cylinder x^2 - y^2 2x can be rewritten through the technique of completing the square. By doing so, we obtain:

x^2 - 2x - y^2 0

Adding and subtracting 1 within the x terms, we get:

x^2 - 2x 1 - 1 - y^2 0

Which simplifies to:

(x - 1)^2 - y^2 1

This equation represents a cylinder that is centered at (1, 0) with a radius of 1.

Step 2: Setting Up the Volume Integral

The volume V of the solid under the paraboloid z x^2 - y^2 and above the xy-plane within the constraints of the cylinder can be expressed as a double integral:

V ?_R (x^2 - y^2) dA

Here, R is the region in the xy-plane bounded by the cylinder.

Step 3: Converting to Polar Coordinates

To simplify the integration, we convert to polar coordinates. The transformation is given by:

x r cos θ, y r sin θ

The Jacobian of the transformation is r. The equation of the paraboloid becomes:

z r^2

Meanwhile, the equation of the cylinder in polar coordinates transforms to:

r^2 cos^2θ - r^2 sin^2θ 2r cosθ, which simplifies to:

r^2 2r cosθ implying r 2 cosθ for r ≠ 0.

Step 4: Determining the Limits of Integration

The region R is defined for r ranging from 0 to 2 cosθ. The angle θ varies from -π/2 to π/2, considering the symmetry of the cylinder about the x-axis.

Step 5: Setting Up the Integral

The volume integral in polar coordinates becomes:

V ∫_{-π/2}^{π/2} ∫_{0}^{2 cosθ} r^2 · r dr dθ ∫_{-π/2}^{π/2} ∫_{0}^{2 cosθ} r^3 dr dθ

Step 6: Evaluating the Inner Integral

First compute the inner integral:

∫_{0}^{2 cosθ} r^3 dr [r^4/4]_{0}^{2 cosθ} (2 cosθ^4)/4 (16 cos^4θ)/4 4 cos^4θ

Step 7: Evaluating the Outer Integral

Now we need to evaluate:

V ∫_{-π/2}^{π/2} 4 cos^4θ dθ

Using the identity for cos^4θ: cos^4θ [1 cos2θ]/2^2 1/4 [1 2 cos2θ cos^2 2θ] Then using cos^2 2θ [1 cos4θ]/2: cos^4θ 1/4 [3/2 2 cos2θ 1/2 cos4θ]

Thus:

V 4 ∫_{-π/2}^{π/2} cos^4θ dθ 4 [3π/2] 3π

Final Answer

The volume of the solid under the paraboloid z x^2 - y^2 and above the xy-plane contained within the cylinder x^2 - y^2 2x is:

boxed{3π}