Hi, everyone.
Welcome back to integralcalc.com.
Today, we’re going to be talking about using the disk and washer method to find the volume
obtained by revolving a surface around an axis.
And in this particular problem, we’ve been told that we’re revolving the surface created
between y equals x squared and x equals y squared.
The surface created between those two curves, we’re going to be revolving that around
the x axis and they’ve asked us to find the volume created by that revolution.
So I’ve gone ahead and graphed the two functions and this is going to be the volume here or
he surface that we revolve around the axis and we’re going to be looking for the volume
obtained by doing that so we’re going to be revolving here around the x axis.
I always want to draw an arrow around the axis about which I am revolving so I can remember
and make sure that I don’t do the wrong axis or something.
So since we’re revolving around the x axis, we need to make sure that both of our functions
are in terms of y.
So notice here that we have a function in terms of y and a function in terms of x.
We need to change the second one so that it’s in terms of y.
If you’re revolving around the y axis, make sure that both of your functions are in terms
of x before you move forward.
So you’ve got the first function which is y minus x squared.
but the second function, x equals y squared we should change into y equals something.
So the way that we’ll do that is by taking the square root of both sides which is going
to give us the square root of x equals y.
So now, we’ve got our two functions in terms of y. y equals x squared and y equals the
square root of x.
And the way that I’ve graphed them here, this curve is y equals x squared.
The upper curve here is y equals square root of x.
So we’ve got our two curves graphed and we’ve been asked to again, revolve around
the x axis.
The formula we’re going to use to find the volume is this formula here that I’ve written
down.
And the first thing you’ll notice that you have limits of integration.
You have a and b.
And you’re usually going to find a and b by setting the two equations equal to one
another.
In this case, I’ll save us the time because it’s basic algebra but what you would do
is essentially set the square root of x equal to x squared.
You’d square both sides.
You’ll get x equals x to the 4th and then you’ll move everything to one side so I’ll
get x to the 4th minus x.
And then I can factor out an x and I can solve.
So I’ll get x cubed minus 1 and you can see that x is going to equal either zero or
1.
So we see pretty quickly here that we’ve got points of intersection here at zero and
also at 1.
So these will be our two intersection points.
And sometimes in a problem like this, you’re given a range on which to revolve and you
can use that.
But in this case, they didn’t give it to us so we had to set them equal to each other,
find the intersection points and you know that those will be our limits of integration.
So we can go ahead and write out our formula here.
We’ll get the volume is equal to the integral from zero to 1.
We’ve got pi here as part of our formula.
Now when it comes to outer radius, remember that we are revolving around the x axis.
This axis here is the axis about which we are going to spin this surface.
So we want to start from the axis of which we’re revolving.
And just go ahead and draw a line perpendicular to the axis up through the surface and up
to the upper function.
y equals the square root of x is above y equals x squared.
So draw a line through the surface all the way up to that.
What we’re going to do here, that line represents the outer radius.
That’s the farthest that the surface is going to be from the axis.
So what we’re going to do is essentially take the square root of x which represents
that outer radius and we’ll put that here into our equation for outer radius.
For the inner radius, since what we’re looking at there is from the axis up to the edge of
the surface before we go through the surface, that will give us up to the line or he graph
of x squared.
so we go ahead and plug in x squared for the inner radius.
And of course we’ve got dx.
So outer, inner, you just go ahead and plug those in.
And then we can go ahead and simplify and solve here.
So what we’ll do is we’ll get pi when we apply that squared there, we’ll get x
minus pi.
According to our formula, the outer radius is squared and the inner radius is squared.
So now, we’ll get x to the 4th dx.
Alright, so we can go ahead and factor out pi out of our integral just to make things
easier.
We’ll get pi times the integral from zero to 1 of x times x to the 4th dx.
And now we can go ahead and integrate.
So volume will be equal to pi times we’ll add 1 to the exponent there and then divide
by the new exponent so 1/2 x squared minus 1/5 x to the 5th and we’ll be evaluating
that on the range zero to 1.
So when we go ahead and plug in 1, of course, we’ll get 1/2 times 1 squared which is just
going to be 1.
And again, 1/5 times 1 to the 5th will just be 1/5.
If we plugged in our lower limit of integration, zero, we would just end up subtracting zero
and zero again here so that’s all going to cancel.
What we’re left with is pi times 1/2 minus 1/5.
If we find a common denominator here, we’ll get 5/10 minus 2/10.
And you can see our final answer is 3/10.
So we’ll get 3/10 pi or you can just write it as 3pi/10.
And that’s it.
That’s our final answer.
So that’s the volume obtained by revolving this surface around the x axis.
So pretty simple.
That’s how you find it.
I hope this video helped you guys and I will see you in the next one.
Bye!