8.1.5How does the volume change?

Changing the Axis of Rotation

You have seen a variety of regions revolved around various lines and axes. Today, you will look at the various volumes formed by rotating the same region in different ways

8-49.

EVERY WHICH WAY

The shaded region shown at right is bounded by the following curves $y = x^2$ , $y = 4$ , and $x = 0$ .

Set up an integral to calculate the volume solid generated when the region is revolved about each line below. Use your graphing calculator to compute the volume.

$x$ -axis

$y$ -axis

$y = -1$

$x = 3$

First quadrant, increasing concave up curve, labeled y = x squared, starting at the origin & ending @ (2, comma 4), with horizontal segment from (0, comma 4) to (2, comma 4), region below segment & above curve shaded.

When calculating volume of a solid, be sure to include the following:

A rough sketch of the solid.
A typical slice with dimensions labeled.
The method used (disk or washers).
The integral that will compute the volume.
The volume found by evaluating the integral.

8-50.

Calculate the volume of each solid generated by the following set of bounded curves revolved about each of the axes below.

$x$ -axis

$y$ -axis

$y = 5$

$x = -1$

$y = -2$

Bounded Curves
$y = 2^x + 1$
$y = 5$
$x = 0$

8-51.

Calculate the volume of the region bounded by $y =\sqrt { x }$ , $y = 0$ , and $x = 4$ revolved about each of the lines below. Use the steps outlined in problem 8-49. Homework Help ✎

$x$ -axis
$y$ -axis
$x = −2$
$y = 5$

First quadrant, increasing concave down curve, labeled y =square root of x, starting at the origin & ending @ (4, comma 2), with vertical segment from (4, comma 0) to (4, comma 2), region below curve, above x axis, & left of x = 4, shaded.

8-52.

Let $y = 3^x + 1$ , $y = 0$ , $x = 0$ , and $x = 2$ . Calculate the volume of the region revolved about each of the lines below. Homework Help ✎

$x$ -axis
$y$ -axis
$y = -1$

8-53.

When setting up a problem to compute volume by the washer method, explain why $\pi \int _ { a } ^ { b } ( R ^ { 2 } - r ^ { 2 } ) d x$ is not equivalent to $\pi \int _ { a } ^ { b } ( R - r ) ^ { 2 } d x$ . Homework Help ✎

Calculus Third Edition