/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 40 Let \(R\) be the region bounded ... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 40

Let \(R\) be the region bounded by the following curves. Use the disk or washer method to find the volume of the solid generated when \(R\) is revolved about the \(y\) -axis. $$y=\sin ^{-1} x, x=0, y=\pi / 4$$

Short Answer

Expert verified
The volume of the solid generated is \(V=\frac{\pi^2}{8}-\frac{\pi}{4}\).

Step by step solution

01

Analyze the given curves

We will first analyze the given equations: - \(y=\sin^{-1}x\): This equation represents the inverse sine function. - \(x=0\): This is a vertical line along the y-axis. - \(y=\pi/4\): This is a horizontal line cutting the inverse sine curve when \(y=\pi/4\). The region R is formed by these equations and when revolved around the y-axis, it forms a solid.
02

Choose the disk or washer method

Since the region R is revolved around the y-axis, we can use the disk method by considering horizontal slices of the region.
03

Set up the equation for a typical disk

The area of a typical disk is \(A(y)=\pi r^2\), where r is the radius. In this case, the radius will be the x-coordinate of the inverse sine function curve, which is \(r=x\). Since \(y=\sin^{-1}x\), we have \(x=\sin y\). Thus, the area of a disk becomes \(A(y)=\pi(\sin y)^2\).
04

Set up the integral to find the volume

To find the volume, we will integrate the area over the given interval for y. The bounds for the integral are from \(y = 0\) to \(y = \pi/4\). Therefore, the integral for the volume will be: $$V = \int_{0}^{\pi/4} \pi(\sin y)^2 dy$$
05

Evaluate the integral

To evaluate the integral, we can use the double-angle formula: \(\sin^2{y} = \frac{1}{2}(1-\cos{2y})\). So, our integral becomes: $$V = \int_{0}^{\pi/4} \pi \cdot \frac{1}{2}(1-\cos{2y}) dy$$ Now, we can integrate: $$V=\pi \cdot \frac{1}{2} \left[ y - \frac{1}{2}\sin{2y} \right]_{0}^{\pi/4}$$ Evaluating at the bounds: $$V = \pi \cdot \frac{1}{2} \left(\left[ \frac{\pi}{4} - \frac{1}{2}\sin{\left(\frac{\pi}{2}\right)}\right] - \left[ 0 - \frac{1}{2}\sin{0} \right]\right)$$
06

Simplify the expression

Simplify the expression, remembering that \(\sin(\pi/2) = 1\) and \(\sin(0)=0\): $$V = \pi \cdot \frac{1}{2} \left( \frac{\pi}{4} - \frac{1}{2} \right)$$
07

Calculate the volume

Calculate the final volume: $$V=\frac{1}{2}\pi\left(\frac{\pi}{4}-\frac{1}{2}\right)=\frac{\pi^2}{8}-\frac{\pi}{4}$$ Thus, the volume of the solid generated when the region R is revolved around the y-axis is \(V=\frac{\pi^2}{8}-\frac{\pi}{4}\).

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