B) Determine the volume of the solid obtained by rotating the region bounded by : y=x -2x and y= x about the line y= 4 Solution First let's get the bounding region and the solid graphed. Again, we are going to be looking for the volume of the walls of this object. Also since we are rotating about a horizontal axis we know that the cross-sectional area will be a function of x. Here are a couple of sketches of the boundaries of the walls of this object as well as a typical ring. The sketch on the left includes the back portion of the object to give a little context to the figure on the right. Now, we 're going to have to be careful here in determining the inner and outer radius as they aren't going to be quite as simple they were in the previous two examples. Let's start with the inner radius as this one is a little clearer. First, the inner radius is NOT x. The distance from the x-axis to the inner edge of the ring is x, but we want the radius and that is the distance from the axis of rotation to the inner edge of the ring. So, we know that the distance from the axis of rotation to the x-axis is 4 and the distance from the x-axis to the inner ring is x. The inner radius must then be the difference between these two. 1) inner radius 2) The outer radius is: 3) The cross-sectional area: A = a [f - g² ] outer inner radius , radius
B) Determine the volume of the solid obtained by rotating the region bounded by : y=x -2x and y= x about the line y= 4 Solution First let's get the bounding region and the solid graphed. Again, we are going to be looking for the volume of the walls of this object. Also since we are rotating about a horizontal axis we know that the cross-sectional area will be a function of x. Here are a couple of sketches of the boundaries of the walls of this object as well as a typical ring. The sketch on the left includes the back portion of the object to give a little context to the figure on the right. Now, we 're going to have to be careful here in determining the inner and outer radius as they aren't going to be quite as simple they were in the previous two examples. Let's start with the inner radius as this one is a little clearer. First, the inner radius is NOT x. The distance from the x-axis to the inner edge of the ring is x, but we want the radius and that is the distance from the axis of rotation to the inner edge of the ring. So, we know that the distance from the axis of rotation to the x-axis is 4 and the distance from the x-axis to the inner ring is x. The inner radius must then be the difference between these two. 1) inner radius 2) The outer radius is: 3) The cross-sectional area: A = a [f - g² ] outer inner radius , radius
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![B)
Determine the volume of the solid obtained by rotating the region bounded by :
y=x -2x and y=x
about the line y- 4
Solution
First let's get the bounding region and the solid graphed.
Again, we are going to be looking for the volume of the walls of this object. Also since we are rotating about a
horizontal axis we know that the cross-sectional area will be a function of x.
Here are a couple of sketches of the boundaries of the walls of this object as well as a typical ring. The sketch
on the left includes the back portion of the object to give a little context to the figure on the right.
Now, we 're going to have to be careful here in determining the inner and outer radius as they aren't going to be
quite as simple they were in the previous two examples.
Let's start with the inner radius as this one is a little clearer. First, the inner radius is NOT x. The distance
from the x-axis to the inner edge of the ring is x, but we want the radius and that is the distance from the axis of
rotation to the inner edge of the ring So, we know that the distance from the axis of rotation to the x-axis is 4
and the distance from the x-axis to the inner ring is x. The inner radius must then be the difference between
these two.
1) inner radius
2) The outer radius is:
3) The cross-sectional aren: A = a [f - g² ]
outer
´inner
radius)
radius,
Using the formulas:
Volume: "-S,a(x)dx
4) Find the volume of this solid: ,](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fca42b389-89df-4627-8846-eef5d3cdb4e1%2F1c383857-060b-44d8-aa82-fb09a5ab6e00%2Fg4x00u6_processed.png&w=3840&q=75)
Transcribed Image Text:B)
Determine the volume of the solid obtained by rotating the region bounded by :
y=x -2x and y=x
about the line y- 4
Solution
First let's get the bounding region and the solid graphed.
Again, we are going to be looking for the volume of the walls of this object. Also since we are rotating about a
horizontal axis we know that the cross-sectional area will be a function of x.
Here are a couple of sketches of the boundaries of the walls of this object as well as a typical ring. The sketch
on the left includes the back portion of the object to give a little context to the figure on the right.
Now, we 're going to have to be careful here in determining the inner and outer radius as they aren't going to be
quite as simple they were in the previous two examples.
Let's start with the inner radius as this one is a little clearer. First, the inner radius is NOT x. The distance
from the x-axis to the inner edge of the ring is x, but we want the radius and that is the distance from the axis of
rotation to the inner edge of the ring So, we know that the distance from the axis of rotation to the x-axis is 4
and the distance from the x-axis to the inner ring is x. The inner radius must then be the difference between
these two.
1) inner radius
2) The outer radius is:
3) The cross-sectional aren: A = a [f - g² ]
outer
´inner
radius)
radius,
Using the formulas:
Volume: "-S,a(x)dx
4) Find the volume of this solid: ,
Transcribed Image Text:Discs and washers: Volume of solids
Determine the volume of the solid obtained by rotating the region bounded by y=x² -4x+5
X= 1, x= 4 and the x and y axes:
Step 1:
The first thing to do is get a sketch of the bounding region and the solid obtnined by rotating the region about the x-
axis. Here are both of these sketches.
Okay, to get a cross section we cut the solid at any x. Below are a couple of sketches showing a typical cross section. The
sketch on the right shows a cut away of the object with a typical cross section without the caps. The sketch on the left
shows just the curve we're rotating as well as its mirror image along the bottom of the solid.
In this case the radius is simply the distance from the x-axis to the curve and this is nothing more than the
function value at that particular x as shown above. The cross-sectional area is then,
A(x) = #(x°- 4x+ 5)° = 7 (x* - 8x² + 26x² - 40x +25)
a) Find the volume of the solid above: V=S a(x) dx
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