Problem 45 - Thomas Calculus

Question

Find the area of the portion of the paraboloid \(x=4-y^{2}-z^{2}\) that lies above the ring \(1 \leq y^{2}+z^{2} \leq 4\) in the \(y z\) -plane.

Step-by-Step Solution

Answer

The area is \( \frac{\pi}{6} \left(17^{3/2} - 5^{3/2}\right) \).

Step 1: Parameterize the Surface

The surface is parameterized using variables \( y \) and \( z \). Since the given equation of the paraboloid is \( x = 4 - y^2 - z^2 \), we can write the position vector of a point \( (x, y, z) \) on the surface as:\[ \vec{r}(y, z) = (4 - y^2 - z^2, y, z) \] where \( y^2 + z^2 \) is within the interval \( [1, 4] \).

Step 2: Compute the Partial Derivatives

Compute the partial derivatives of \( \vec{r}(y,z) \) with respect to \( y \) and \( z \):- \( \frac{\partial \vec{r}}{\partial y} = (-2y, 1, 0) \)- \( \frac{\partial \vec{r}}{\partial z} = (-2z, 0, 1) \).

Step 3: Find the Cross Product

Determine the cross product of the partial derivatives to find the normal vector:\[ \frac{\partial \vec{r}}{\partial y} \times \frac{\partial \vec{r}}{\partial z} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -2y & 1 & 0 \ -2z & 0 & 1 \end{vmatrix} = \hat{i} \times (1, 0) - \hat{j} \times (-2y, 0) + \hat{k} \times (-2y, -2z) \]This simplifies to:\[ (1, 2z, 2y) \].

Step 4: Compute the Magnitude of the Cross Product

Find the magnitude of the cross product vector, which is:\[ \| (1, 2z, 2y) \| = \sqrt{1^2 + (2z)^2 + (2y)^2} = \sqrt{1 + 4z^2 + 4y^2} \].

Step 5: Set Up the Double Integral

Set up the double integral to compute the surface area over the region \( 1 \leq y^2 + z^2 \leq 4 \):\[ \iint_{D} \sqrt{1 + 4z^2 + 4y^2} \; dy \, dz \]where \( D \) is the annular region in the \( yz \)-plane defined by the given ring.

Step 6: Convert to Polar Coordinates

To simplify the integration, convert to polar coordinates where \( y = r \cos \theta \) and \( z = r \sin \theta \):- \( dy \, dz = r \, dr \, d\theta \)Update the limits of integration for \( r \) from \( 1 \) to \( 2 \) and \( \theta \) from \( 0 \) to \( 2\pi \).

Step 7: Evaluate the Integral

Evaluate:\[ \int_0^{2\pi} \int_1^2 \sqrt{1 + 4r^2} \, r \, dr \, d\theta \]Perform the inner integral first:\[ \int_1^2 r \sqrt{1 + 4r^2} \, dr \]Make a substitution, let \( u = 1 + 4r^2 \), \( du = 8r \, dr \).Thus, it updates to \( \frac{1}{8} \int 2 du \rightarrow \frac{1}{8} u^{3/2} \mid_{} \) with proper bounds from \( u(1) = 5 \) to \( u(2) = 17 \).Therefore, evaluate:\[ \frac{1}{8} [\frac{2}{3} (17^{3/2} - 5^{3/2})] \] and multiply by \( \int_0^{2\pi} d\theta = 2\pi \).

Step 8: Calculate Final Result

After integration calculations, obtain:\[ A = 2\pi \times \frac{1}{8} \times \frac{2}{3} \left(17^{3/2} - 5^{3/2}\right) \]Solve this expression to get the numerical value for the surface area.

Key Concepts

Parameterization of Surfaces

When working with surfaces in calculus, parameterization helps us express the surface using two variables instead of three. For the paraboloid described by the equation \( x = 4 - y^2 - z^2 \), the surface can be parameterized with the components \( y \) and \( z \).
This means expressing every point on the surface as a vector function \( \vec{r}(y, z) = (4 - y^2 - z^2, y, z) \).
This process is crucial as it allows the use of calculus techniques to analyze and compute properties like the surface area.

• The parameterization provides a continuous map from the plane, described by \( y \) and \( z \), to the surface.
• This representation simplifies the computation of derivatives and integrals over the surface.

Hence, it's a stepping-stone for performing integration over a surface, leading us to the desired measurements.

Polar Coordinates

Polar coordinates are particularly useful in simplifying double integrals over circular or annular regions. Instead of using \( y \) and \( z \) directly, we express them in terms of \( r \) and \( \theta \):
\( y = r \cos \theta \) and \( z = r \sin \theta \).
Here, \( r \) represents the distance from the origin, and \( \theta \) denotes the angle from the positive \( y \)-axis. This transformation redefines the integral's limits to represent the given annulus region in a simpler form.

• The annular region's bounds, \( 1 \leq y^2 + z^2 \leq 4 \), translate into \( 1 \leq r \leq 2 \).
• The integration limits for \( \theta \) span from 0 to \( 2\pi \), covering a full circle.

By utilizing polar coordinates, the double integral setup becomes more manageable and often easier to work with.

Double Integral

The double integral in calculus is a technique to compute the surface area or volume over a two-dimensional region. For the paraboloid, calculating the area involves setting up a double integral over the annular region in the \( yz \)-plane.
The integral is represented as:
\( \iint_{D} \sqrt{1 + 4z^2 + 4y^2} \, dy \, dz \).
Here, \( D \) is the domain defined by the annular region.

This expression gives the surface area element of the paraboloid.

• By converting it to polar coordinates and setting \( dy \, dz = r \, dr \, d\theta \), the integral simplifies further.
• This conversion makes it feasible to handle the complex integral limits smoothly over defined intervals for \( r \) and \( \theta \).

Hence, the double integral is a powerful approach to determine the size characteristics of surfaces or shapes.

Calculus

In calculus, understanding curves and surfaces is fundamental. Calculus provides the tools to measure things like areas and volumes, especially when dealing with complex shapes like paraboloids.
The problem-solving method incorporates several key calculus concepts:

• **Differentiation**: To find the partial derivatives, \( \frac{\partial \vec{r}}{\partial y} \) and \( \frac{\partial \vec{r}}{\partial z} \), which are needed to establish the tangent plane.
• **Vector Cross Product**: Used to find the normal vector, crucial in determining the area of a surface element.
• **Integration**: Both single and double integrals help compute entire areas or volumes by summing up infinitely small elements.

A combined approach using all these elements aids in solving complex geometrical problems efficiently.
Whether finding the surface area of parts of the paraboloid or other 3D shapes, calculus methods provide the necessary analysis and computations.