Engineering Mathematics Exam > Engineering Mathematics Videos > Shell method around a non-axis line (Part - 2) - Mathematics
Shell method around a non-axis line (Part - 2) - Mathematics Video Lecture - Engineering Mathematics
FAQs on Shell method around a non-axis line (Part - 2) - Mathematics Video Lecture - Engineering Mathematics
| 1. What is the shell method in mathematics? |  |
Ans. The shell method is a technique used in calculus to find the volume of a solid of revolution. It involves integrating the product of the circumference of a shell (a cylindrical slice) and its height to determine the volume.
| 2. How is the shell method different from the disk method? | |
Ans. The shell method and the disk method are both used to find the volume of solids of revolution, but they differ in the orientation of the slices used. The shell method uses cylindrical shells aligned parallel to the axis of rotation, while the disk method uses disks or washers perpendicular to the axis of rotation.
| 3. Can the shell method be applied to solids of revolution around a non-axis line? | |
Ans. Yes, the shell method can be applied to solids of revolution around a non-axis line. In such cases, the shells are still cylindrical slices, but their orientation and size change depending on the non-axis line. The key is to carefully determine the height and circumference of each shell based on the given non-axis line.
| 4. What are some applications of the shell method in engineering? | |
Ans. The shell method has various applications in engineering. It is commonly used to calculate the volume of objects with rotational symmetry, such as pipes, tanks, and cylinders. Additionally, it can be applied in the analysis of structures with curved surfaces, such as beams and arches, to determine their strength and stability.
| 5. Are there any limitations to using the shell method for volume calculations? | |
Ans. While the shell method is a powerful tool for calculating volumes of solids of revolution, it does have some limitations. It can be challenging to apply when dealing with irregular shapes or objects with multiple non-axis lines. In such cases, alternative methods like the disk method or integration by slicing may be more suitable. Additionally, the shell method assumes a uniform thickness of the shells, which may not always be the case in real-world engineering scenarios.
Text Transcript from Video
Where I left off, we were just
essentially chugging through
this fairly hairy derivative--
this definite integral--
this antiderivative.
It takes my brain a
little while to come
up with the next term.
So we evaluated at 2 and
now we have to subtract
this evaluated at 1.
So minus 16 pi minus 4
pi over 3-- oh, sorry.
Plus.
The minus is going to
be on everything.
Oh no, sorry.
That is a minus.
That's a minus.
Plus 4 pi.
Right?
Because x is 1.
Minus pi over 2.
And now we can simplify it.
Let's see what we can do.
This is really a hairy problem.
The 16 pi minus 8 pi,
that equals 8 pi.
And then that's a plus.
The 32 plus 8 pi.
32 pi plus 8 pi,
that equals 40 pi.
So let's see, I've simplified
it to 40 pi, and what's minus
8 times 4 is 32 pi over 3.
And then all of that, let's
see if I can simplify this.
Let's see, 16 pi plus
4 pi, that's 20 pi.
And then, let's see.
Minus 4 pi over 3
minus pi over 2.
So let's get a common
denominator for this
part right here.
So if I put everything over 6--
20 pi over 6 is the same thing
is 120 pi over 6, and then
minus-- 4 pi over 3, if I put
it over 6 it becomes
8 pi over 6, right?
And then pi over 2, if I put it
over 6 it becomes 3 pi over 6.
So minus 3 pi.
So this whole term that we're
going to have to subtract from
this term is equal to
120 minus 11, right?
So that's 109 pi over 6.
This equals 1-- is that right?
Yeah, 109 pi over 6.
See what happens when you
make up problems on the fly?
You get ugly numbers.
109 pi over 6.
And what does that
top part translate?
So this is what we're
going to subtract.
This is when we evaluated
the antiderivative at 1.
And let's simplify this one.
So that one, if we put 3 common
denominator, that's 120 pi
over 3 minus 32 pi over 3.
120 minus 32, let's
see, we get 90-- 88?
Right.
88 pi over 3.
So that equals 88 pi over
3, and remember, that's
just the top part.
And then-- this is more of a
review of fractions than
anything else-- and then if
I want to put it over
6, I just double it.
So I think we're almost done.
Let me switch colors.
So if you go over a
denominator of 6-- 88
pi over 3-- let's see.
If I double that, I
get 160-- 176, right?
176 pi minus this.
Minus 109 pi.
I'm sure I made a
careless mistake.
These are my least favorite
type of things to do.
Hairy fractions.
So 176 minus 109.
That's the same thing
as 76 minus 9.
And that is 65.
So our final answer
is 65 pi over 6.
Which isn't as hairy
as it seemed when we
started the problem.
But that's pretty neat.
This is kind of a
strange shape.
It's kind of a wide ring that
has-- the upper part of the
ring and the inner part of
the ring are hard edges.
But then it's curved
on the outside.
And we were able to figure
out the volume of that.
Especially-- and what was weird
about this, is when it was
rotated around the line
y is equal to minus 2.
Hopefully I didn't confuse you.
These are about as difficult
as these volume of
revolution problems get.
If you want more,
just let me know.
I will see you in
the next video.
essentially chugging through
this fairly hairy derivative--
this definite integral--
this antiderivative.
It takes my brain a
little while to come
up with the next term.
So we evaluated at 2 and
now we have to subtract
this evaluated at 1.
So minus 16 pi minus 4
pi over 3-- oh, sorry.
Plus.
The minus is going to
be on everything.
Oh no, sorry.
That is a minus.
That's a minus.
Plus 4 pi.
Right?
Because x is 1.
Minus pi over 2.
And now we can simplify it.
Let's see what we can do.
This is really a hairy problem.
The 16 pi minus 8 pi,
that equals 8 pi.
And then that's a plus.
The 32 plus 8 pi.
32 pi plus 8 pi,
that equals 40 pi.
So let's see, I've simplified
it to 40 pi, and what's minus
8 times 4 is 32 pi over 3.
And then all of that, let's
see if I can simplify this.
Let's see, 16 pi plus
4 pi, that's 20 pi.
And then, let's see.
Minus 4 pi over 3
minus pi over 2.
So let's get a common
denominator for this
part right here.
So if I put everything over 6--
20 pi over 6 is the same thing
is 120 pi over 6, and then
minus-- 4 pi over 3, if I put
it over 6 it becomes
8 pi over 6, right?
And then pi over 2, if I put it
over 6 it becomes 3 pi over 6.
So minus 3 pi.
So this whole term that we're
going to have to subtract from
this term is equal to
120 minus 11, right?
So that's 109 pi over 6.
This equals 1-- is that right?
Yeah, 109 pi over 6.
See what happens when you
make up problems on the fly?
You get ugly numbers.
109 pi over 6.
And what does that
top part translate?
So this is what we're
going to subtract.
This is when we evaluated
the antiderivative at 1.
And let's simplify this one.
So that one, if we put 3 common
denominator, that's 120 pi
over 3 minus 32 pi over 3.
120 minus 32, let's
see, we get 90-- 88?
Right.
88 pi over 3.
So that equals 88 pi over
3, and remember, that's
just the top part.
And then-- this is more of a
review of fractions than
anything else-- and then if
I want to put it over
6, I just double it.
So I think we're almost done.
Let me switch colors.
So if you go over a
denominator of 6-- 88
pi over 3-- let's see.
If I double that, I
get 160-- 176, right?
176 pi minus this.
Minus 109 pi.
I'm sure I made a
careless mistake.
These are my least favorite
type of things to do.
Hairy fractions.
So 176 minus 109.
That's the same thing
as 76 minus 9.
And that is 65.
So our final answer
is 65 pi over 6.
Which isn't as hairy
as it seemed when we
started the problem.
But that's pretty neat.
This is kind of a
strange shape.
It's kind of a wide ring that
has-- the upper part of the
ring and the inner part of
the ring are hard edges.
But then it's curved
on the outside.
And we were able to figure
out the volume of that.
Especially-- and what was weird
about this, is when it was
rotated around the line
y is equal to minus 2.
Hopefully I didn't confuse you.
These are about as difficult
as these volume of
revolution problems get.
If you want more,
just let me know.
I will see you in
the next video.
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Oct 27, 2024 Last updated |
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