Hi everyone.
Welcome back to integralCALC.com.
Today we’re going to be doing another volumes by cylindrical shells problem and in this
problem, we’re given two functions y squared equals x and x equals 2y and they’ve asked
us to actually find the area in between these two functions when both functions are rotated
around the y-axis.
So, a couple of interesting things to note with this problem.
First of all, we’re going to be using the same formula that we always use for volumes
by cylindrical shells when we’re rotating around the y-axis and everything is the same
here.
You will notice that we have the x here in the formula which indicates that we’re rotating
around the y-axis and we’re going to have our function here and that’s going to be
something interesting to pay attention to.
But oftentimes with a problem like this, we will be given a range on which we’re evaluating
from a to b and we use those values to plug in here for our limits of integration a and
b.
But in this case, we’re not given a and b.
So the first thing that we need to do is find them ourselves and then once we do, we can
plug them in, set up our integral and we will be able to simplify and solve.
So the first thing we want to do is find a and b like I said and the way that we’re
going to do that is we’re going to solve both of these equations for y and then set
them equal to one another.
So if you look at the first equation here, to solve this first equation y squared equals
x for y, we will want to take the square root of both sides to make this exponent here go
away.
So when we take the square root of both sides, taking the square root of y squared will just
simply give us y and then we will be left with the square root of x over here on this
side.
So you can see that that’s what we have over here, y equals the square root of x.
Solving this equation for y, if we just divide both sides by 2, that will also leave y on
its own so we’re left with y equals x divided by 2 and you can see we’ve done that here
with our second equation.
So now that we’ve solved both of them for y, now that we know that the square root of
x is equal to y and that x over 2 is equal to y, we can set the square root of x equal
to x over 2 and simplify this equation here to solve for x and this will actually give
us the points at which these two functions intersect one another.
So we solve this function for x or this equation for x by first multiplying both sides by 2.
I think this is the easiest way.
So we multiply both sides by 2 and then we divide both sides by the square root of x.
Now remember that the square root of x is the same thing as x to the one-half so you
can see from here, that I divided both sides by the square root of x and I just changed
it automatically to x to the one-half instead of the square root of x because they’re
the same thing but it will make the next step easier.
So we go ahead and divide by the square root of x and now notice that we have – this
is basically x to the 1, right?
That 1 is implied as an exponent.
x to the 1 over x to the one-half and whenever you have a fraction and you have the same
base in the numerator and the denominator, in our case our base is x, you can subtract
the exponent in the denominator from the exponent in the numerator.
So we’re going to subtract one-half, this exponent from 1 which is the implied exponent
in the numerator.
So we will have 1 minus one-half is one-half so we end up with – this right side here
simplifies to x to the one-half.
So we’re left with that and now, we want to get x by itself so we go ahead and we square
both sides because if we square x to the one-half, we will end up multiplying one-half by 2 and
the 2s will cancel and we will be left with x to the 1, in other words just x, and that’s
how we can solve for x.
We square both sides.
One-half times 2 equals 1 so we get x to the 1 over here equals 4.
So we found the point at which our two functions intersect with one another but remember, we
need two points at which they intersect.
So we know they intersect at 4 but we need another point and if we go back up here and
we look, you can see zero is usually a good bet to try, like zero or 1 especially if you’re
in calculus 2, if you’ve got 4, zero or 1; and if we look up here and we plug in zero
for x to both equations, we plug in zero here, we will get y equals the square root of zero.
In other words, y equals zero and if we plug in zero to this equation, we will get y equals
zero for 2 and of course that’s y equals zero.
So we can prove that zero is also a point at which they intersect.
Another way that you can look at that is by graphing the two equations here.
So if you graph the two equations, you will see you have y equals the square root of x
in purple and y equals x over 2 in orange.
Here’s our point of intersection at 4.
You can also see that they intersect at zero.
So if you’re not sure, you can graph them on your calculator and find the points of
intersection or verify them that way.
So we know that our functions intersect at zero and 4 so we’ve gone ahead and we plugged
in the limits of integration, zero and 4.
Now we leave 2 pi x because that’s part of our formula here but we have to plug in
for f of x and normally, in a volumes by cylindrical shells problem, we’re simply given one function
and rotate it around the y-axis between x equals 1 and x equals 4 and we have this one
function and we plug in our one function in place of f of x in our formula.
In this case, they’ve asked for the area between the two graphs.
So what we have to do is we need to graph them and look at which one is higher or above
the other one.
In this case, you can see that y equals the square root of x and purple is above y equals
x over 2 in orange.
You can see that it’s higher than or above the other graph.
So what we need to do is take the function that’s higher which is the square root of
x and subtract from that the lower function which is x over 2.
So instead of just plugging in one function here for f of x, we take the higher function
and subtract from that the lower function and now give us the area in between the two
graphs.
If you don’t have access to a graphing calculator and you can’t graph them, you’ve already
found your limits of integration zero and 4.
Just pick any point between zero and 4 – one, two, or three for example.
Plug them in to both of your equations and see which one is higher and that will be the
greater function so you will know that that one here goes first and then you subtract
the lower function.
So that’s another way to do it.
So now we’ve set up our integral and we’re going to go ahead and start simplifying this
so that we can integrate.
The first thing we’re going to do is distribute this x so we end up with x times the square
root of x minus x times x over 2 and now we’re going to go ahead and multiply these together.
Again, here x times the square root of x, we have like bases of x and x.
This first one here is x to the 1 power and the second one here is square root of x is
x to the one-half power.
So when we have like bases and they’re multiplied together, we add the exponents so we have
1 here plus one-half is one and a half or three-halves.
So we end up with x to the three-halves power.
This term obviously we have an x squared over 2.
I’ve gone ahead and separated the coefficient which is the one-half from the x squared to
make it a little bit simpler.
So that’s about as far as we can simplify our integral.
We also move to the 2 pi out in front because it’s a constant and it’s a coefficient
on all of these variables so we can just bring it out in front of the integral to save ourselves
some hassle.
So now we’re going to go ahead and integrate and what that looks like of course, we will
add 1 to three-halves and we get five-halves and then we divide the coefficient on this
x term which is 1.
It’s implied and not written but it’s a 1.
We divide the 1 by the new exponent which is five-halves.
So for this coefficient, we end up with 1 divided by five-halves which 1 divided by
any fraction is going to simplify to the inverse of that fraction so we end up with a two-fifths
out here in this coefficient and then same thing here.
We add one to the exponent and we get 3 and then we divide our coefficient one-half by
the new exponent 3 and we get one-sixth.
So we have one-sixth here, x to the third and we’re going to be evaluating between
these limits of integration zero and 4.
So let’s go ahead and evaluate this definite integral here.
Remember with a definite integral when we’re evaluating, we always plug in the top number
which was 4 to our whole function and then always subtract here in the middle and then
plug in our bottom number which you will remember was zero.
So we plug that in to our function as well and now it’s going to be a lot of arithmetic
to simplify but what we will end up with in the front here, this simplifies to sixty four-fifths.
This simplifies to thirty two-thirds and then of course you see we have zero multiplied
here in this first term and zero in the second term so this whole thing here is going to
become zero so we will end up with 2 pi times zero.
When we simplify one more step, obviously the second term goes away completely.
Sixty four-fifths minus thirty two-thirds, we find the common denominator of 15 and we
end up with thirty two-fifteenths.
When we multiply that by 2 pi, we get our final answer which is 64 pi over 15.
So what this tells us is that the area when we rotate these two functions here, both of
them around the y-axis, the area that’s created between those two curves is 64 pi
over 15.
So I hope that video helped you guys and I will see you in the next one.
Bye.
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