Problem 49 - Thomas Calculus

Question

Find the area of the surfaces in Exercises \(49-54\) $$ \begin{array}{l}{\text { The surface cut from the bottom of the paraboloid } z=x^{2}+y^{2} \text { by }} \\ {\text { the plane } z=3}\end{array} $$

Step-by-Step Solution

Answer

The area of the surface is \( 6\pi \).

Step 1: Understand the Problem

We are tasked with finding the area of the surface beneath the plane \( z = 3 \) and above the paraboloid \( z = x^2 + y^2 \). This implies we find the surface cut out by these two equations.

Step 2: Setup the Projection onto the XY-Plane

Since \( z = x^2 + y^2 \) is a paraboloid, it will intersect the plane \( z = 3 \) to form a circle on both the surface of the paraboloid and the plane. The projection onto the xy-plane is given by setting \( z = 3 \) in \( z = x^2 + y^2 \), so, \( x^2 + y^2 = 3 \).

Step 3: Parameterize the Surface

The portion of the paraboloid under the plane can be parameterized using polar coordinates where \( x = r\cos(\theta) \) and \( y = r\sin(\theta) \) with the condition \( 0 \leq r \leq \sqrt{3} \) and \( 0 \leq \theta \leq 2\pi \).

Step 4: Express the Surface in Terms of Parameters

Parameterizing \( z = x^2 + y^2 \), we can express \( z \) in terms of \( r \) as \( z = r^2 \). Therefore, the surface is parameterized by \( \mathbf{r}(r, \theta) = (r\cos\theta, r\sin\theta, r^2) \).

Step 5: Compute Surface Area Element

The surface area \( A \) can be evaluated using the formula \( A = \int\int \| \mathbf{r}_r \times \mathbf{r}_\theta \| \, dr \, d\theta \), where \( \mathbf{r}_r \) and \( \mathbf{r}_\theta \) are the partial derivatives of \( \mathbf{r} \) with respect to \( r \) and \( \theta \) respectively.

Step 6: Calculate Partial Derivatives

Find \( \mathbf{r}_r = (\cos\theta, \sin\theta, 2r) \) and \( \mathbf{r}_\theta = (-r\sin\theta, r\cos\theta, 0) \).

Step 7: Evaluate the Cross Product

Calculate \( \mathbf{r}_r \times \mathbf{r}_\theta = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \cos\theta & \sin\theta & 2r \ -r\sin\theta & r\cos\theta & 0 \end{vmatrix} = (-2r^2\cos\theta, -2r^2\sin\theta, r) \).

Step 8: Find the Magnitude of the Cross Product

Compute \( \| \mathbf{r}_r \times \mathbf{r}_\theta \| = \sqrt{(-2r^2\cos\theta)^2 + (-2r^2\sin\theta)^2 + r^2} = r\sqrt{4r^2 + 1} \).

Step 9: Set Up the Integral for Surface Area

Substitute \( \| \mathbf{r}_r \times \mathbf{r}_\theta \| = r\sqrt{4r^2 + 1} \) into the area integral: \( A = \int_{0}^{2\pi} \int_{0}^{\sqrt{3}} r\sqrt{4r^2 + 1} \, dr \, d\theta \).

Step 10: Evaluate the Integral

First, integrate with respect to \( r \): \( \int_{0}^{\sqrt{3}} r\sqrt{4r^2 + 1} \, dr \), and then integrate the resultant value with respect to \( \theta \) from \( 0 \) to \( 2\pi \).

Step 11: Calculate Indefinite Integral

Using substitution, let \( u = 4r^2 + 1 \), \( du = 8r \, dr \), so \( \, dr = \frac{1}{8r} \, du \), the integral becomes \( \frac{1}{8} \cdot \frac{1}{2} \cdot \sqrt{u} \, du \), which resolves to \( \frac{1}{12}(u^{3/2}) \). Evaluate from \( r = 0 \) to \( r = \sqrt{3} \).

Step 12: Evaluate Integral and Multiply by Angular Part

Calculate: \[ \frac{1}{12} [ (4(\sqrt{3})^2 + 1)^{3/2} - 1 ] \]. Compute this value and multiply it by \( 2\pi \) to find the total surface area. The result is \( 6\pi \).

Key Concepts

Paraboloid

A paraboloid is a 3-dimensional surface that represents a quadratic equation, specifically in the form \( z = x^2 + y^2 \). This type of surface resembles a bowl or a parabola translated into three-dimensional space. In our problem, the paraboloid opens upwards, meaning it extends higher as you move away from its center at the origin on the xy-plane.
This type of geometric shape has some interesting characteristics:

• Symmetry: It is symmetrical about its axis, in this case, the z-axis.
• Shape: It forms an elliptical shape when intersected with a plane parallel to its base.
• Vertex: The lowest point, or the vertex, of the paraboloid when it opens upwards is located at the origin (0,0,0).

Understanding the structure of a paraboloid can help in visualizing areas where certain planes, like our plane \( z = 3 \), can intersect the paraboloid, creating shapes like circles.

Polar Coordinates

Polar coordinates are used to represent points in a two-dimensional space using a radius and an angle, making it especially convenient for circular or radial symmetries. In the given exercise, polar coordinates help to easily describe the circle formed by the intersection of the paraboloid and plane.
Here's a quick breakdown:

• The point is determined by \( (r, \theta) \), where \( r \) is the distance from the origin, and \( \theta \) is the angle measured from the positive x-axis.
• The transformation from cartesian to polar coordinates can be expressed as \( x = r\cos(\theta) \) and \( y = r\sin(\theta) \).

This system simplifies calculations and parameterizations by utilizing symmetries. For example, the circle \( x^2 + y^2 = 3 \) can be easily represented in polar coordinates as a constant radius \( r = \sqrt{3} \). This transformation is instrumental when parameterizing surfaces or computing areas that possess radial symmetries.

Partial Derivatives

Partial derivatives measure how a function changes as one of its variables changes, while keeping all other variables constant. These derivatives are crucial in multivariable calculus when dealing with surfaces and their properties.
In the context of our problem:

• We use partial derivatives to find the surface's slope or rate of change in the \( r \) and \( \theta \) directions, given the parameterization \( \mathbf{r}(r, \theta) = (r\cos(\theta), r\sin(\theta), r^2) \).
• The partial derivatives \( \mathbf{r}_r \) and \( \mathbf{r}_\theta \) provide the tangent vectors needed to form a basis for the tangent plane at any point on the surface.

By computing these vectors, we derive essential information about the orientation and composition of the surface, which is necessary for calculating the area of a complex 3D shape.

Cross Product

The cross product is a vector operation that returns a vector perpendicular to two given vectors in three-dimensional space. It is vital in calculating areas of surfaces, especially in vector calculus.
Here’s how it applies to our task:

• A cross product of the partial derivatives \( \mathbf{r}_r \) and \( \mathbf{r}_\theta \) produces a normal vector to the surface defined by our parameterization.
• This normal vector has a magnitude that corresponds to the area of the parallelogram formed by \( \mathbf{r}_r \) and \( \mathbf{r}_\theta \).

The magnitude \( \| \mathbf{r}_r \times \mathbf{r}_\theta \| \) is pivotal as it gives us the differential area element when integrating over the entire surface. Understanding how to compute and utilize cross products is fundamental in handling surface integrals and hence finding the total surface area.