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Studying for a test? Prepare with these 8 lessons on Applications of definite integrals.

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# Shell method for rotating around horizontal line

Video transcript

What we're going to do in this
video is take the function y is equal to the cube
root of x and then rotate this around the x-axis. And if we do that, we
get a solid of revolution that looks like that. And we're doing it between x is
equal to 0 and x is equal to 8. And you get something
that looks like this. And you could find the
volume of this actually quite easily using
the disk method. But just to show you that you
could do it an alternate way, we're going to use
the shell method. But we're going to use
the shell method now to rotate around a horizontal
line, and specifically the x-axis. So how would we do this? Well, what we want to do,
what you could imagine, is constructing a rectangle
that looks like this. Let me do it in
this salmon color. So you have a rectangle that
looks something like that. Its depth or its height,
you could say, is dy. And then its length
right over here is going to be 8 minus
whatever x value this is. Let me make this clear. The width is going to
be 8 minus whatever x value this is right over here. And you might already realize
that if this is going to be dy, we're going to be integrating
with respect to y over an interval of y. And so we really want to have
everything in terms of y. So this x value, whatever
it is, as a function of y. So if y is equal to
the cube root of x, we can cube both
sides, and we can get x is equal to y
to the third power. These are equivalent
statements right over here. And so, this distance right over
here is going to be 8 minus y to the third, if we were to
express it in terms of y. And then when you rotate
that thing around the x-axis, it's going to construct
the outside of the cylinder or shell, as we like to call it. And I'll try my best
to draw that shell. So it will look
something like this. So you're going to have a
shell that looks like this. And there you go. And actually, this shares
a common boundary right over here. So you're going to have a shell
that looks something like that. So that hopefully
helps a little bit. Let me give it some depth. You'll have a shell that
looks something like that. If we could figure out
the volume of that shell, which is really going to be
the area of the outer surface area times the depth. And then if we were to sum
up all of those shells, this shell is for a
particular y in our interval. If we were to sum up over all
of the y's in our interval, all the volumes of
the shells, then we have the volume of this figure. So once again, how do we figure
out the volume of a shell? Well, we can figure
out the circumference of I guess the left or the
right, in this case-- the left or the right of our cylinder. If we can figure out
that circumference, that circumference is equal
to 2 pi times the radius. And what is the radius? The radius of these things are
just going to be your y value. That's this distance
right over here. That's just going
to be the y value. So it's going to be
equal to 2 pi times y. And then if we want the
area of the outer surface, we just multiply
the circumference times the width of our cylinder. So let me write it over here. So outer surface area is
going to be equal to our 2 pi y times our 8 minus
y to the third. 8 minus y to the third
is just this length. You multiply that
times circumference. You get the outer surface area. Now, if you want to find the
volume of this one shell, it's going to be the
outer surface area. 2 pi y times 8 minus y to
the third times the depth, times this dy right over here. I'll do the dy in purple. So that's the
volume of one shell. If we want to find the volume of
the entire solid of revolution, we have to sum all these up
and then take the limit as they become infinitely
thin, and we have an infinite number
of these shells. So we're going to take the
sum from, and remember, we're dealing in y. So the volume is
going to be equal to, so what's our interval
in terms of y? So y definitely starts off at
0, and when x is equal to 8, what is y? Well, 8 to the 1/3
power is just 2. So y is 2, this value
right over here. Let me make it a
little bit clearer. This value right over here is 2. So y goes from 0 to 2, and
we've set up our integral. And this one looks
pretty straightforward. So I think we can crank
through it in this video. So this is going to be equal
to, we can take out the 2 pi, 2 pi times the definite
integral from 0 to 2. Let's multiply the y of
8y minus y to the fourth. All of that dy. This is going to be equal to
2 pi times the anti-derivative of this business. Anti-derivative of
8y is 4y squared. Anti-derivative of 1
negative y to the fourth is negative y to
the fifth over 5. And we're going to
evaluate it at 0 and 2. So this is going to be
equal to 2 pi times, we evaluate this business at 2,
2 squared is 4 times 4 is 16. 16 minus 2 to the fifth is 32. So minus 32/5. And then, you evaluate this
stuff at 0, you just get the 0. So that's what we're left with. And now, we just
have to simplify this thing a little bit. So let's see. 16 over 5, this part
right over here. 16 is the same thing,
I should say, as 80/5. And from that, we
are subtracting 32/5. And so, that is equal to 48/5. So all of this business
is equal to 48/5. Did I do that right? 80 minus 30 is 50, and then
minus another 2 gets us to 48. So this is 48/5 times 2 pi. And now we deserve
our drum roll. 48 times 2 is 96. Let me do this in
a new color, just to emphasize that
we're at the end. 48 times 2 is 96
times pi over 5. So once again, this
is something that you could have solved using the
disk method in terms of x. And we're just showing that
you could also solve it in the shell method
in terms of y. The shell or the hollowed
out cylinder method, whatever you want to
call it, in terms of y.