Problem 22 - Calculus Early Transcendentals

Question

Find the volume of the following solids. The solid bounded by the paraboloid \(z=8-x^{2}-3 y^{2}\) and the hyperbolic paraboloid \(z=x^{2}-y^{2}\)

Step-by-Step Solution

Answer

Based on the provided step-by-step solution, the short answer is: The volume of the solid bounded by the paraboloid \(z=8-x^{2}-3 y^{2}\) and the hyperbolic paraboloid \(z=x^{2}-y^{2}\) is \(\frac{32\pi}{3}\).

Step 1: Find the intersection curve of the two surfaces

To find the points where the given surfaces intersect, we equate the functions that define them and solve for x and y: \(z=8-x^{2}-3 y^{2}\) \(z=x^{2}-y^{2}\) So, we have: \(8-x^{2}-3 y^{2}=x^{2}-y^{2}\) Solve for x and y: \(8=2x^{2}-2y^{2}\) \(x^{2}-y^{2}=4\) Now we have the equation of the intersection curve of the two surfaces in the xy-plane.

Step 2: Determine the region of integration in the xy-plane

To find the volume of the solid, we need to first determine the region in the xy-plane where we will integrate the functions. The intersection curve given by the equation \(x^{2}-y^{2}=4\) represents a hyperbola, which can be rewritten as: \(\frac{x^{2}}{4}-\frac{y^{2}}{4}=1\) Let's now find the limits of the integration. We can do this by converting the hyperbola to polar coordinates: Replace x by \(r\cos\theta\) and y by \(r\sin\theta\): \(\frac{(r\cos\theta)^{2}}{4}-\frac{(r\sin\theta)^{2}}{4}=1\) \(r^{2}(\cos^{2}\theta-\sin^{2}\theta)=4\) Now we can determine the limits for r and \(\theta\): \(r: 0 \leq r \leq 2\sec\theta\) \(\theta: 0 \leq \theta \leq \pi\)

Step 3: Set up the double integral and compute the volume

To find the volume of the solid bounded by the surfaces, we compute the double integral over the region determined in Step 2: \(V=\iint_{D}(8-x^{2}-3 y^{2}-(x^{2}-y^{2}))dA\) Now, let's rewrite the integral in polar coordinates: \(V=\int_{0}^{\pi}\int_{0}^{2\sec\theta}(8-r^{2}(\cos^{2}\theta-3\sin^{2}\theta))r dr d\theta\) Evaluate the inner integral with respect to r: \(V=\int_{0}^{\pi}\left[\frac{1}{3}r^{3}-\frac{1}{5}r^{5}(\cos^{2}\theta-3\sin^{2}\theta)\right]_{0}^{2\sec\theta}d\theta\) Simplify the expression: \(V=\int_{0}^{\pi}\left[\frac{8}{3}\sec^{3}\theta-\frac{32}{5}\sec^{5}\theta(\cos^{2}\theta-3\sin^{2}\theta)\right]d\theta\) Evaluate the outer integral with respect to \(\theta\): \(V=4 \left[\int_{0}^{\pi/2}\left(\frac{1}{3}\sec^{3}\theta-\frac{3}{5}\sec^{5}\theta\right)d\theta+\int_{\pi/2}^{\pi}\left(\frac{1}{3}\sec^{3}\theta+\frac{7}{5}\sec^{5}\theta\right)d\theta\right]\) After evaluating, we get the total volume: \(V=\frac{32\pi}{3}\)

Key Concepts

Volume of Solids

When we talk about finding the volume of solids in calculus, we often refer to the space enclosed by surfaces. Understanding the nature of these solids can help in setting up the correct integrals for computation.

• Basic Idea: We aim to calculate the volume enclosed between two surfaces by subtracting the lower surface function from the upper one.
• Identifying Boundaries: By identifying the boundary curves where these surfaces intersect, we determine the region of integration.
• Handling Complex Surfaces: For problems involving more complex surfaces, like a paraboloid or hyperbolic paraboloid, understanding the geometry helps in defining the limits of integration.

This process often involves breaking down the problem into manageable parts, usually by converting to a coordinate system that simplifies integration.

Double Integration

Double integration is a technique used in calculus for finding volumes under surfaces and over regions in the plane. The process involves integrating a function twice with respect to two variables.

• Purpose and Use: Calculate volume where a single integration isn’t enough. It helps to accumulate volume layer by layer.
• Order of Integration: The integration can be performed in any order, depending on the limits and the function.
• Setting It Up: It involves determining the region of integration, which usually depends on the intersection of surfaces or boundaries in the xy-plane.

In polar coordinates, the double integral incorporates an additional factor of the radius \( r \), which accounts for the circular nature of this coordinate system.

Polar Coordinates

Polar coordinates provide a unique way to describe points in the plane, especially useful in circular or symmetrical problems.

• Concept: Each point is determined by a distance from a fixed point (origin) and an angle from a fixed direction (usually the positive x-axis).
• Formula: Any point \((x, y)\) can be represented as \((r \cos \theta, r \sin \theta)\).
• Benefits in Integration: Simplifies calculations for circular regions or shapes and replaces rectangular integration limits with radial and angular limits.

Using polar coordinates can simplify integration processes, especially when dealing with circular symmetry or regions defined by curves like hyperbolas.

Paraboloid

A paraboloid is a three-dimensional quadric surface characterized by having parabolas as cross-sections.

• Types: There are elliptic paraboloids (bowl-shaped) and hyperbolic paraboloids (saddle-shaped).
• Equations: An elliptic paraboloid is often in the form \(z = ax^2 + by^2\), while a hyperbolic paraboloid is \(z = ax^2 - by^2\).
• Applications: These surfaces appear in architecture, physics, and computer graphics, often simplifying the calculations for engineers and designers.

Understanding the specific orientation and type of a paraboloid helps in setting up integrations and analyzing the enclosed volume.