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Calculating integral with shell method - Video Lecture - Engineering
FAQs on Calculating integral with shell method - Mathematics
| 1. What is the shell method in calculus? |  |
Ans. The shell method is a technique used in calculus to calculate the volume of a solid of revolution. It involves integrating the product of the circumference of a cylindrical shell and its height over a given interval.
| 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 shape of the elements used for integration. The shell method uses cylindrical shells, while the disk method uses circular disks or washers.
| 3. How do you set up the integral for the shell method? | |
Ans. To set up the integral for the shell method, you need to express the height and radius of the cylindrical shells in terms of the variable of integration. The integral is then constructed by integrating the product of the circumference of the shell and its height over the desired interval.
| 4. Can the shell method be used to find the volume of any solid of revolution? | |
Ans. The shell method can be used to find the volume of solids of revolution that can be generated by rotating a curve around a vertical or horizontal axis. However, if the curve intersects the axis of rotation, the method may need to be modified or a combination of methods may be required.
| 5. Are there any limitations or drawbacks to using the shell method? | |
Ans. One limitation of the shell method is that it can be more complex to set up the integral compared to the disk method, especially for shapes that require multiple integrals or have varying radii. Additionally, the shell method may not be applicable for certain irregular or asymmetric shapes that do not lend themselves well to cylindrical shells.
Text Transcript from Video
In the last video we
were able to set up
this definite integral using
the shell or the hollow cylinder
method in order to
figure out the volume
of this solid of revolution.
And so now let's just
evaluate this thing.
And really the main
thing we have to do here
is just to multiply
what we have here out.
So multiply this expression out.
So this is going to be equal
to-- I'll take the 2 pi out
of the integral.
2 pi times the
integral from 0 to 1.
Let's see, 2 times
the square root
of x is 2-- I'll write it
as 2 square roots of x.
But I'll write it
as 2x to the 1/2.
It'll make it a
little bit easier
to take the antiderivative
conceptually, or at least
in our brain.
So two times the square
root of x is 2x to the 1/2.
2 times negative x squared
is negative 2 x squared.
And then we have negative x
times the square root of x.
Well, that's x to the
first times x to the 1/2.
That's going to be negative
x to the 3/2 power.
And then we have negative
x times negative x
squared that's going to be
positive x to the third power.
And all of that dx.
And so now we're ready to
take the antiderivative.
So this is going to be equal to
2 pi times the antiderivative
of all of this business
evaluated at 1 and at 0.
So the antiderivative
of 2 times x to the 1/2
is going to be 2-- it's
going to be-- let's see.
We're going to take x
to the 3/2 times 2/3.
So it's going to be
4/3 x to the 3/2.
And then for this
term right over here
it's going to be negative
2/3 x to the third.
And you could take
the derivative here
to verify that you
actually do get this.
And then right over here, let's
see, if we incremented this,
you get x to the 5/2.
And so we're going to
want to multiply by 2/5.
So minus-- let me do
this in another color.
Let's see, so this one
right over here, it's
going to be minus 2/5
x to the 5/2 power.
Yep, that works out.
And then finally you're going
to have x to the fourth over 4
plus-- let me do that
in a different color--
plus x to the fourth over 4.
That's this term
right over here.
And now we just have
to evaluate at 1 and 0.
And 0, luckily, all of these
terms end up being a 0.
So that's nice and cancels out.
And so we are just
left with-- we're
just-- [INAUDIBLE] cancel out.
It just evaluates to 0.
So this is just 2 pi times when
you evaluate all this business
at 1.
So that's going to be 4/3
minus 2/3 minus 2/5 plus 1/4.
And the least common
multiple right over here
looks like 60, so
we're going to want
to put all this over
a denominator of 60.
So it's going to be 2 pi
times all of this business
over a denominator of 60.
And 4/3 is same thing as 80/60.
Negative 2/3 is the same
thing as negative 40/60.
Negative 2/5 is the same
thing as negative 24/60.
And then 1/4 is the
same thing as 15/60.
So this is equal to--
and actually this
will cancel over
here, and you'll just
get a 30 in your denominator.
So in your denominator,
you get a 30.
And up here 80 minus 40 is 40.
40 minus 24 gets us to 16.
16 plus 15 is 31.
So we get 31 times pi
over 30 for the volume
of the figure right over there.
were able to set up
this definite integral using
the shell or the hollow cylinder
method in order to
figure out the volume
of this solid of revolution.
And so now let's just
evaluate this thing.
And really the main
thing we have to do here
is just to multiply
what we have here out.
So multiply this expression out.
So this is going to be equal
to-- I'll take the 2 pi out
of the integral.
2 pi times the
integral from 0 to 1.
Let's see, 2 times
the square root
of x is 2-- I'll write it
as 2 square roots of x.
But I'll write it
as 2x to the 1/2.
It'll make it a
little bit easier
to take the antiderivative
conceptually, or at least
in our brain.
So two times the square
root of x is 2x to the 1/2.
2 times negative x squared
is negative 2 x squared.
And then we have negative x
times the square root of x.
Well, that's x to the
first times x to the 1/2.
That's going to be negative
x to the 3/2 power.
And then we have negative
x times negative x
squared that's going to be
positive x to the third power.
And all of that dx.
And so now we're ready to
take the antiderivative.
So this is going to be equal to
2 pi times the antiderivative
of all of this business
evaluated at 1 and at 0.
So the antiderivative
of 2 times x to the 1/2
is going to be 2-- it's
going to be-- let's see.
We're going to take x
to the 3/2 times 2/3.
So it's going to be
4/3 x to the 3/2.
And then for this
term right over here
it's going to be negative
2/3 x to the third.
And you could take
the derivative here
to verify that you
actually do get this.
And then right over here, let's
see, if we incremented this,
you get x to the 5/2.
And so we're going to
want to multiply by 2/5.
So minus-- let me do
this in another color.
Let's see, so this one
right over here, it's
going to be minus 2/5
x to the 5/2 power.
Yep, that works out.
And then finally you're going
to have x to the fourth over 4
plus-- let me do that
in a different color--
plus x to the fourth over 4.
That's this term
right over here.
And now we just have
to evaluate at 1 and 0.
And 0, luckily, all of these
terms end up being a 0.
So that's nice and cancels out.
And so we are just
left with-- we're
just-- [INAUDIBLE] cancel out.
It just evaluates to 0.
So this is just 2 pi times when
you evaluate all this business
at 1.
So that's going to be 4/3
minus 2/3 minus 2/5 plus 1/4.
And the least common
multiple right over here
looks like 60, so
we're going to want
to put all this over
a denominator of 60.
So it's going to be 2 pi
times all of this business
over a denominator of 60.
And 4/3 is same thing as 80/60.
Negative 2/3 is the same
thing as negative 40/60.
Negative 2/5 is the same
thing as negative 24/60.
And then 1/4 is the
same thing as 15/60.
So this is equal to--
and actually this
will cancel over
here, and you'll just
get a 30 in your denominator.
So in your denominator,
you get a 30.
And up here 80 minus 40 is 40.
40 minus 24 gets us to 16.
16 plus 15 is 31.
So we get 31 times pi
over 30 for the volume
of the figure right over there.
About this Video
Apr 20, 2026 Last updated
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