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Chapter 9: Problem 48

Use the integration capabilities of a graphing utility to approximate to two decimal places the area of the surface formed by revolving the curve about the polar axis. $$ r=\theta, \quad 0 \leq \theta \leq \pi $$

Short Answer

Expert verified
The answer will be the numerical result obtained from the graphing utility, rounded to two decimal places. The specific answer cannot be provided in this format as a graphing utility is needed to calculate it.

Step by step solution

01

Identify the formula

For a curve given in polar coordinates, the surface area \( A \) of the surface formed by revolving the curve from \( \theta=a \) to \( \theta=b \) about the polar axis is given by the formula: \[A = 2\pi \int_a^b r \sin(\theta) d\theta\]Here, \( r = \theta \) and the limits of integration are from 0 to \( \pi \).
02

Substitute the function and the limits into the formula

The formula becomes:\[A = 2\pi \int_0^\pi \theta \sin(\theta) d\theta\]This is the integral that needs to be evaluated.
03

Evaluate the integral using a graphing utility

To evaluate the integral using a graphing utility, input the function \( \theta \sin(\theta) \) and find the definite integral from 0 to \( \pi \). Round the numerical result to two decimal places.

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The initial and terminal points of a vector \(v\) are given. (a) Sketch the directed line segment, (b) find the component form of the vector, and (c) sketch the vector with its initial point at the origin. Initial point: (2,-1,-2) Terminal point: (-4,3,7)

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