Today we’re going to be doing a disk and washer problem.
And in this particular example, we’ve been asked to find the area of the region bounded
by 1 – x^2 and y = 0 when that region is revolved around the y axis.
With these disk and washer method problems, the first thing I like to do is draw a picture
if I can and sometimes that can be harder than other times depending on the curves that
you’re dealing with.
In this case, it’s not too difficult because 1 – x^2 and y = 0 are not difficult functions
to graph.
So let’s go ahead and graph a picture of what we’ve got going on over here.
If we go ahead and draw a picture of these two curves, we’ll call this one here y = 1
– x^2, we’ll do that in orange.
If we say that this is 1 along the y axis and this is 1 along the x axis, a picture
of y = 1 – x^2 of the graph would look like this, an upside down parabola or a parabola
that points down.
That’s the picture of that one and then if we call y = 0 in green, of course that’s
easy.
That’s just right along the x axis here in green.
As you can see, you know that we’re going to rotate around the y axis so we’re going
to do a rotation like that and we have this region here that’s bounded by these two
curves.
What we’re going to do, because we’re asked to rotate around the y axis, we’re
going to take a cross-section that’s perpendicular to the y axis so you can almost imagine a
cross section of this being a little rectangular piece of our curve like that and basically
when we revolve this around the y axis, what we’re looking at is half of a sphere that’s
sitting on top of the x axis.
So if we drew it out in 3-D, it would look like this with the y axis coming here and
the x axis coming here and it’s sitting right on top of the x axis.
When we rotate it around, we’re just going to get a little sliver of this half-sphere
and we’re going to find the area based on that kind of image.
So because we’re rotating around the y axis, one of the first things we’re going to need
to do is make sure that we solve both of our functions for x if we can.
We have y equals something for both of our functions.
We need to change them both into x equals something if possible.
Instead of y = 1 – x^2, let’s go ahead and add x^2 to both sides.
We’ll get x^2 + y = 1 and then again we want to solve for x so we’ll go ahead and
subtract y from both sides and get x^2 = 1 – y and then to solve for x, we’ll take
the square root of both sides and we’ll get x equals the square root of 1 – y.
So that’s our first function.
Our second function y = 0, we can’t solve in terms of x because it’s just a horizontal
line.
There’s nothing we can do about that but it doesn’t really matter because we know
that our limits of integration are going to be zero and 1.
Normally, when we wrote it around the x axis which is what we do more commonly, we’re
looking at limits of integration like this from zero to 1 but because we’re rotating
around the y axis, we’re going to be looking at limits of integration along the y axis.
So the first one being 0 and the second one being 1.
We know our limits of integration, we have our functions in terms of x, we know when
we’re talking disk-washer method here that there is no washer.
This is just going to be a disk because we have this line here along the x axis, y = 0
and we’re rotating along the y axis.
There’s not going to be a hole here in the middle.
If we were integrating from 1/2 to 1, we would
have a hole in the middle.
But because we’re integrating from 0 to 1, there is no hole in the graph.
There’s no washer, it’s just a disk.
That means our limits of integration are 0 and 1 so we can start putting together our
formula for the volume of this rotated region.
You know our limits of integration are from zero to one along the y axis.
We know that our outer function is this orange function here.
We’ve solved for it in terms of x so we’re actually going to use this one down here x
equals the square root of 1 – y^2.
We’ll call that the formula for a rotated region like this is volume equals limits of
integration from a to b pi times the outer curve squared minus pi times the inner curve
squared dx if you’re rotating around the x axis.
In our case, we’re going to have pi.
That’s part of the formula.
Our outer curve is going to be the square root of 1 – y that we solve for, square
that.
And because there’s no washer here, because there’s nothing to subtract here, there’s
no hole, we can leave this part out.
And then instead of dx, because we’re rotating around the y axis, we’re going to call this
dy and we’re going to integrate with respect to y.
Now this becomes fairly easy.
We’re just going to simplify the integral as much as possible before we integrate.
The square of the square root of 1 – y is just 1 – y dy.
We can pull pi out in front of the integral since it’s a constant and it’s a coefficient
here on this y – 1 term so we end up with just 1 – y.
And then when we integrate, the integral of 1 with respect to y is simply y.
The integral of –y here is -1/2 y^2 and then we’re going to be evaluating on the
range 0 to 1.
Now we can just go ahead and first plug in 1 so we’ll get 1 – 1/2 times 1^2 = 1/2
then we have to subtract and plug in our lower limits of integration.
Our limit of integration is zero so we get zero minus 1/2 times 0^2 is still zero.
So now that we have plugged in our limits of integration, the zeros are going to go
away and we’ll be left with 1 – 1/2 which is 1/2 and when we simplify that, we’ll
get 1/2 pi or pi/2.
And that is the volume of this region rotated around the y axis.
So again, the volume bounded by the x axis here or the function y = 0 and this curve
here rotated around the y axis, the volume of that region is pi/2.
So that’s it.
I hope this video helped you guys and I will see you in the next one.