BAM! Mr. Tarrou. In this lesson we are going to learn an alternative way or another way
of finding the volume of a solid of revolution. This is going to compliment what we already
know which is finding these volumes using disk method, the washer method and the shell
method. And using the theorem of Pappus would be an option when, well either you are required
or you find it relatively easy to find the centroid of a plane region. Okay...so if you
can find that center of mass in a plane region then this is an option that you can use. Let
R be a region in a plane and let L be a line in the same plane such that line L does not
intersect the interior of R, which this sort of region in a plane or just region. If r
is the distance between the centroid and that is the center of mass for a planar lamina
or just a thin plate, of R or our plane region and line L then the volume V of the solid
of revolution formed by revolving R about this line is- our volume is going to be equal
to 2 pi r times A which is the area of our region. Now, our two examples are not going
to be based off of this diagram. I am just using this to lay down the variables for us.
We have this plane region which of course in this example is a rectangle. We have our
centroid marked by x bar and y bar, we have r which is that distance between our line
of rotation and our centroid- in this case it would be a horizontal distance or a distance
of x and we have our line- now this being vertical axis of rotation, this might be the
y axis or x=5 or something like that. And while we need to find the centroid, in this
particular diagram with our axis of rotation being vertical, all we would actually need
to figure out which is going to be (x ) ̅ and of course (x ) ̅ is the moment about
the y axis divided by m. Now we might need the y ̅ as well but with the way this diagram
is set up, that is the part of the centroid that would be very important to being able
to figure out this volume. SO let’s get to our first of two examples. BAM! Example
1: Use the Theorem of Pappus to find the volume of the solid of revolution, the torus formed
by revolving the circle x^2+〖(y+5)〗^2=9 about the x axis. Okay, well of course by
now we know how to graph a circle. We have a circle whose center is at (0, -5) and whose
radius is the √9 or 3. So we have a circle with a center at (0, -5), a radius of 3 and
we are revolving this again about the x axis. So I have done the best I can, basically try
and draw a 3 dimensional doughnut if you will and... Excuse me. And this problem took really
honestly longer to draw than it is going to be to solve. The second example will be a
lot more interesting. So as we look at our theorem of Pappus, it's 2 pi r, now... okay
now that was the distance between the axis of rotation or the line L and our centroid.
Well our centroid for the circle, our center of mass, I won’t need to do any fancy kind
of calculations to figure that out like I will in the next example. It’s simply the
center of the circle so it’s going to be (0,-5) so the distance between the x axis
and -5 of course is a distance of 5, so r is 5 times the area of our plane region. Well
again we don’t need to set up a definite integral to find the area of a circle, that's
simple pi r^2. And our r value is 3. So simply cleaning this up, we have 3 squared is 9 and
9 times 5 is 45 and 45 times 2 is equal to 90 and we have pi times pi here, so that going
to be 90 pi^2 and we are talking about volume, so you might want to say, its units cubed.
And just to double check, that is absolutely correct. Well that didn’t take very long.
Let’s get on to our second and last example that requires a lot more calculus to figure
out the problem. Second example and our last, Use the Theorem of Pappus to find the volume
of the solid of revolution formed by revolving the region bounded by the graphs of y= square
root of x, y=0 and x=4 and the line we are revolving it around the line of x=6. So what
information do we need? Well we need to know what r is, the distance between the line and
the centroid the center of mass and we need to know the area of the region of the plane
or the plane region. Okay so we have this diagram set up to match this example and if
it sounds familiar if you watch all my calculus videos, it will sound familiar. We have already
found the volume of the solid of revolution two ways- the Washer method and the Shell
method and now because we know how to find the centroid or the center of mass of a planar
lamina, we can now use, well our theorem of Pappus. So what so we need? Well we need r,
the distance between the axis of rotation our line L and our centroid. Well since our
axis of rotation is vertical, we need a horizontal distance. So, though we are told and we need
to find the centroid or the center of mass of this two dimensional plane basically, we
really only care about the (x ) ̅, the x coordinate of the center of mass for this
particular problem. So again just a reminder (x ) ̅ is = m sub y, the moment about the
y axis divided by m or the mass of the planar lamina and that is equal to p times the definite
integral from a to b of x times [f(x)-g(x)] dx. And p is our uniform density of our planar
lamina or our thin plate or plane. And for this example, they are going to cancel out
anyway. The mass is going to be p times definite integral of [f(x)-g(x)] dx and really that’s
just going to be for these problems of finding these planar laminas, really just going to
be the area of that plane region. So we need a moment about the y axis and the mass of
the planar lamina. Okay so let’s move this over here and try to work out some of this
so we can finish the problem. Okay let’s go ahead and find the area of that plane region.
Well the area, remember we were talking about this graph, this plane region bounded by y=
square root of x, x=4 and y=0. SO that is going to be equal to, I don’t
need the P’s, cause they are going to cancel once we set up this formula here. Maybe I
should just go ahead and do that. So the area is going to be the definite integral from
0 to 4 of f(x) - g(x) well that’s f(x) - g(x), it’s being bound by the x axis, so it’s
just going to be y= square root of x or x to the ½ power minus y=0 and of course writing
a subtraction of zero doesn’t do anything so we can just say that it is the definite
integral from 0 to 4 of x^ ½ dx Well of course we are going to say that- that is now equal
to, raising that power by 1 and then dividing by 3/2 gives us a multiplication of 2/3 and
so we have definite integral from 0 to 4 of x^ ½ dx becomes 2/3 x to the 3/2 power with
our upper limit of 4 and our lower limit of 0. And that is going to be 2/3 times 4 to
the 3/2 power. That is going to be 2/3... The square root of 4 is equal to 2 of course,
that raised to the 3, now we have 2/3 times 8 and that is going to be 16/3. So that is
the area of this plane region. So let’s write this over here.
Excellent! And that’s also going to be if we let that p out front, that would be the
mass of our planar lamina. Now the moment about the y axis. Well that is going to be
equal to, and again the p are going to cancel out, definite integral from a to b of x times
[f(x)-g(x)] and again f(x) is the square root of x minus g(x) which again is that lower
bound of y=0, so x^1/2 minus 0. And that is going to give us... Oh our lower and upper
limits again are between 0 and 4. And let’s reveal this process one step at a time to
shorten up the video. So working through our power rule, increasing that by 1, dividing
by 5/2 which is multiplying by 2/5 and plugging in our upper and lower limits of 0 and 4,
we have our moment about the y axis which is equal to 64 /5. Alright so now that those
calculations are done, lets, again we need to know 2 pi r times the area. What’s the
area? 16/3. We don’t know what r is yet. We don’t know that distance between our
axis of rotation or line L and the x coordinate of our centroid because that all we need for
this particular problem. SO we need to finish that up. Well (x ) ̅ is equal to the moment
about the y axis which is 64/5 divided by the mass which because these p's are cancelling
out, really it’s just the area in this particular case. 16/3. And that’s comes out to be 64/5
times 3/16. 16 divided by 16 is 1 and 64 divided by 16 is 4 and we 12 let’s try that again
12/5. So (x ) ̅ is 12/5. Well if the x coordinate of our centroid is 12/5 and our line L or
our axis of rotation is 6, then we need to figure out... well what’s that distance?
That’s going to be from referencing the y axis, that’s going to be a distance of
6 minus (x ) ̅. So r, therefore, is going to be, well the greater value, I am going
to write 12/5 but of course the greater x value is 6, you can’t have a negative distance
minus 12/5 is going to be equal to, putting that 6 over 1, finding a common denominator,
we get- the fact that I am running out of room here, sorry about that-. 6 times 5 is
30. 30 minus 12 or 30 minus 10 is 20 minus 2 is 18. 18 over 5. Alright. Now we have everything
we need. The volume of this solid is going to be equal to 2 pi times r which is 18/5
times the area of our plane region which is 16/3 is equal to. Now I don’t want to try
and do 2 times 18 times 16 off the top off my head. So off the top of my head, we get-
the final answer is equal to 576 pi. That’s not what I said. Let’s try that again. 576.
Okay I am glad that this lesson is almost over. Let’s try this again. 576 pi over
15 and that reduces to 192/5 pi and that is the volume of this solid which we have now
found using the washer method, using the shell method and even though I stopped being able
to copy correctly at the end of the video, we have found the correct answer using the
Theorem of Pappus. I am Mr. Tarrou. BAM! Go do your homework.