Problem 49 - Thomas Calculus

Question

Find the area of the surfaces. The surface cut from the bottom of the paraboloid \(z=x^{2}+y^{2}\) by the plane \(z=3\)

Step-by-Step Solution

Answer

The area of the surface is \( 2\pi (2 - \sqrt{2}) \).

Step 1: Understand the Problem

We need to find the area of the surface that is created when the paraboloid given by the equation \( z = x^2 + y^2 \) is cut by the plane \( z = 3 \). The surface lies below this plane.

Step 2: Identify Bounds for Integration

Since the surface is cut by the plane \( z = 3 \), we know that \( x^2 + y^2 = 3 \). This represents a circle of radius \( \sqrt{3} \) centered at the origin in the \( xy \)-plane.

Step 3: Set Up the Integral

The surface area integral for a surface \( z = f(x, y) \) is given by:\[\text{Area} = \iint_{D} \sqrt{1 + \left( \frac{\partial z}{\partial x} \right)^2 + \left( \frac{\partial z}{\partial y} \right)^2} \, dx \, dy\]For \( f(x, y) = x^2 + y^2 \), calculate the partial derivatives: \[ \frac{\partial f}{\partial x} = 2x, \quad \frac{\partial f}{\partial y} = 2y \]

Step 4: Compute Integrand

Substitute the partial derivatives in, so the integrand becomes:\[\sqrt{1 + (2x)^2 + (2y)^2} = \sqrt{1 + 4x^2 + 4y^2}\]

Step 5: Switch to Polar Coordinates

We will switch to polar coordinates where \( x = r \cos\theta \), \( y = r \sin\theta \), and \( dx \, dy = r \, dr \, d\theta \). Then: \\( x^2 + y^2 = r^2 \) and so:\[\sqrt{1 + 4r^2}\]The region of integration becomes \( 0 \leq r \leq \sqrt{3} \) and \( 0 \leq \theta \leq 2\pi \).

Step 6: Evaluate the Integral

Set up the integral in polar coordinates:\[\int_{0}^{2\pi} \int_{0}^{\sqrt{3}} \sqrt{1 + 4r^2} \cdot r \, dr \, d\theta\]First, evaluate the integral with respect to \( r \), and then evaluate with respect to \( \theta \).* The integral with respect to \( r \):\[ \int_{0}^{\sqrt{3}} r \sqrt{1 + 4r^2} \, dr \]* Use substitution \( u = 1 + 4r^2, \ du = 8r \, dr \), leading to:\[ \frac{1}{8} \int \sqrt{u} \, du \]* Evaluate this integral and then use the limits.* Integrate the result with respect to \( \theta \) from 0 to \(2\pi\).

Step 7: Simplify and Final Result

After integration, the result simplifies to:\[ \text{Area} = \pi \left( \frac{1}{6} \cdot (4(1 + 4(\sqrt{3})^2)^{3/2} - 1) \right) \] where you evaluate and simplify to get the final numerical result.

Key Concepts

Surface Integration

Surface integration is a method used to find the area of surfaces in three-dimensional space. Essentially, it involves summing up tiny pieces of a surface to determine its total area. It relies on calculus concepts since the surface must be expressed as a function of two variables.

When dealing with surface integration, the integrand often involves the derivatives of the function that defines the surface. For instance, if you have a surface defined by the equation \( z = f(x, y) \), the surface area \( A \) is calculated by:

• First, finding the partial derivatives \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \).
• Substitute these into the surface integral formula \[ \text{Area} = \iint_{D} \sqrt{1 + \left( \frac{\partial z}{\partial x} \right)^2 + \left( \frac{\partial z}{\partial y} \right)^2} \, dx \, dy \].

Thus, this integral calculates the "bent" area of the surface rather than just a "flat" projected area.

In the given problem, knowing how to correctly set up and compute surface integrals is essential for finding the area of the paraboloid sliced by a plane.

Polar Coordinates

Polar coordinates provide a more intuitive way to solve integrals over circular regions or when dealing with problems that have radial symmetry. They simplify complex Cartesian coordinates to those involving radius and angle.

In polar coordinates, the positions of points are described by:

• Radius \( r \), the distance from the origin.
• Angle \( \theta \), the angle from the positive x-axis.

To convert from Cartesian coordinates \( (x, y) \) to polar coordinates, use:

• \( x = r \cos\theta \)
• \( y = r \sin\theta \)
• The Jacobian determinant for changing variables gives \( dx \, dy = r \, dr \, d\theta \).

For the paraboloid problem, integration is performed over a circular domain. Thus, converting the integral to polar coordinates simplifies the calculations, especially the bound setup. The choice of limits of integration, with \( 0 \leq r \leq \sqrt{3} \) and \( 0 \leq \theta \leq 2\pi \), reflects the circular nature of the region in question.

Plane Cutting Paraboloid

A paraboloid defined by \( z = x^2 + y^2 \) is a 3-dimensional surface shaped like an upward-opening bowl. When a plane, such as \( z = 3 \), intersects a paraboloid, the intersection forms a shape on the xy-plane, in this case, a circle.

For this problem, the vertical plane slices through the paraboloid at \( z = 3 \). Substituting this value into the paraboloid equation shows where the cut occurs:

• \( x^2 + y^2 = 3 \) forms a circle of radius \( \sqrt{3} \).

This circle is essentially the boundary for the surface area to be calculated. The plane effectively "caps" the paraboloid, creating a finite surface to work with in the integration.

Understanding the geometry of how a plane cuts a 3D shape such as a paraboloid is vital when determining various surface areas and helps form correct integration bounds in problems involving such intersections.