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Chapter 5: Problem 20

(a) Find the area of the surface \(z=1+x^{2}+y^{2}\) inside the cylinder \(x^{2}+y^{2}=1\). (b) Find the volume inside the cylinder between the surface and the \((x, y)\) plane. Use cylindrical coordinates.

Short Answer

Expert verified
Area \(= \frac{\pi}{6} (5^{3/2} - 1) \), Volume \(= \frac{3\pi}{2} \).

Step by step solution

01

Understand the problem

(a) Area of surface under z = 1 + x² + y² inside the cylinder x² + y² = 1. (b) Volume inside the cylinder x² + y² = 1, between the surface and (x, y) plane.
02

Convert to cylindrical coordinates

In cylindrical coordinates, use: \[ x = r \cos \theta, \; y = r \sin \theta, \; z = z \]Here, \( x^2 + y^2 = r^2 \). So the surface becomes \( z = 1 + r^2 \) and the boundary is \( r = 1 \).
03

Compute surface area integral

The formula for the surface area is \[ A = \iint_{D} \sqrt{1+\left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2} \; dA \]Converting to cylindrical coordinates and simplifying:\[ \iint_{D} \sqrt{1 + 4r^2} \; r \; dr \; d\theta \]The limits are from 0 to 2π for \( \theta \) and 0 to 1 for \( r \).Therefore, the integral becomes: \[ \int_{0}^{2\pi} \int_{0}^{1} \sqrt{1+4r^2} \; r \; dr \; d\theta \]
04

Perform the integral

Solve the integral for \( r \): \[ \int_{0}^{1} r \sqrt{1+4r^2} \; dr \]Set \( u = 1 + 4r^2 \), hence \( du = 8r \; dr \) and \( dr = \frac{du}{8r} \), yielding:\[ \int \frac{1}{8} u^{1/2} du \]Thus, it becomes: \[ \frac{1}{12} \left[ u^{3/2} \right]_{1}^{5} = \frac{1}{12} \left[ (1+ 4 \times 1)^{3/2} - 1\right] = \frac{1}{12} (5^{3/2} - 1) \]Multiply this by the integral over \( \theta \) to get the surface area.\[ 2\pi \times \frac{1}{12} \left( 5^{3/2} - 1 \right) \]
05

Compute volume integral

The volume inside the cylinder, bounded by \( z = 1 + r^2 \) and the plane, is given by:\[ V = \iint_{R} \ (1+r^2) \, r \, dr \, d\theta \]Following similar limits, we get:\[ V = \int_{0}^{2\pi} \int_{0}^{1} (1+r^2) \, r \, dr \, d\theta \]
06

Perform the volume integral

Separate the integral into two parts:\[ \int_{0}^{1} r \, dr + \int_{0}^{1} r^3 \, dr \]Which evaluates to: \[ \left. \frac{r^2}{2} \right|_{0}^{1} + \left. \frac{r^4}{4} \right|_{0}^{1} = \frac{1}{2} + \frac{1}{4} = \frac{3}{4} \]Then multiplying by the \( \theta \) integral:\[ 2\pi \times \frac{3}{4} = \frac{3\pi}{2} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

cylindrical coordinates
Cylindrical coordinates are a three-dimensional coordinate system that extends polar coordinates by adding a height dimension. Instead of using Cartesian coordinates \( x \), \( y \), and \( z \), cylindrical coordinates are represented as \( (r, \theta, z) \). Here, \( r \) is the radial distance from the origin in the \( xy \)-plane, \( \theta \) is the angular position around the \( z \)-axis, and \( z \) is the height above the \( xy \)-plane.
Using cylindrical coordinates can significantly simplify certain types of integrals, especially when dealing with problems that have radial symmetry. In our example, we convert \( x \) and \( y \) to \( r \) and \( \theta \) for easier calculation. We have:
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Most popular questions from this chapter

(a) Revolve the curve \(y=x^{-1}\), from \(x=1\) to \(x=\infty\), about the \(x\) axis to create a surface and a volume. Write integrals for the surface area and the volume. Find the volume, and show that the surface area is infinite. Hint: 'The surface area integral is not easy to evaluate, but you can casily show that it is greater than \(\int_{1}^{\infty} x^{-1} d x\) which you can evaluate. (b) 'The following question is a challenge to your ability to fit together your mathematical calculations and physical facts: In (a) you found a finite volume and an infinite area. Suppose you fill the finite volume with a finite amount of paint and then pour off the excess leaving what sticks to the surface. Apparently you have painted an infinite area with a finite amount of paint! What is wrong? (Compare Problem \(15.31 \mathrm{c}\) of Chapter 1.)

(a) Write a triple integral in spherical coordinates for the volume inside the cone \(z^{2}=x^{2}+y^{2}\) and between the planes \(z=1\) and \(z=2\). Evaluate the integral. (b) Do (a) in cylindrical coordinates.

Find the area cut from a spherical surface of radius \(a\) by a square hole of side \(a \sqrt{2}\), whose axis is along a diameter of the sphere. Himt for evaluating the integral: Change to polar coordinates and evaluate the \(r\) integral. Then use the following formula: $$ \int_{0}^{k / 4} \sqrt{2-\sec ^{2} \theta} d \theta=\frac{\pi}{2}(\sqrt{2}-1) $$

Find the volume in the first octant bounded by the coordinate planes and the planc \(x+2 y+z=4\)

Find the area of the part of the cone \(2 z^{2}=x^{2}+y^{2}\) in the first octant cut out by the planes \(y=0\), and \(y=x / \sqrt{3}\), and the cylinder \(x^{2}+y^{2}=4\).
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