Hi, everyone.
Welcome back to integralcalc.com.
Today, we’re going to be doing another Green’s theorem problem.
And in this particular problem, we’ve been given two functions, p of x,y and q of x,y
with p of x,y equal to y plus e to the x and q of x,y equal to 2x squared plus cos y.
We’ve also been given this line integral.
it’s defined by the three corners of a triangle. (0,0), (1,1), and (2,0) are the vertices of
the triangle and they define the triangle here on the first quadrant of our graph.
So we’ve gone ahead and graphed them.
What we’ve been asked to do is find the area bounded by this line integral.
And keep in mind that when you’re doing a Green’s theorem problem, you’re usually
given the problem in one of two forms.
You’re either given the two functions, p of x,y and q of x,y as we’ve been given
in this problem or you might be given something that’s like this line integral here with
the two functions already plugged in.
And if that’s the case, it would look like this.
You would have the functions.
What you want to do is recognize p of x,y here and q here.
So p remember is in front of dx and q will be in front of dy.
So either way, whether you’re given the line integral or the functions themselves,
you can extract the functions from the line integral so that you can identify which one
is p, which one is q, and then start taking your partial derivatives.
But remember that Green’s theorem allows us to convert this line integral into a double
integral so that we can evaluate the area.
So this formula here for the double integral is what we’re going to be focusing on.
And notice that to use this double integral, we’re going to need the partial derivative
of q with respect to x and the partial derivative of p with respect to y.
We’ll also need to fill in upper and lower limits of integration on both of these integrals
here and replace these da with dy,dx.
But we’ll get there in a second.
So the first thing we’re going to do is take the partial derivative of q with respect
to x and the partial derivative of p with respect to y.
So when we take the partial derivative of q with respect to x, remember that we’re
treating y as a constant and x is the variable.
So q is 2x squared plus cosine of y.
So with respect to x, that gives us 4x plus zero which of course is just 4x.
That cosine of y becomes zero because remember y is a constant.
When we take the partial derivative of p with respect to y, again we treat x as the constant
and y as the variable this time.
So we’ll get 1 plus zero which of course just gives us 1.
So now that we have these, we can go ahead and plug in and we’ll use the same colors
here so we’ll use blue for the partial derivative of q with respect to x.
We’re going to plug it in the formula this way.
And the partial derivative of p with respect to y is going to plug into our integral here.
So let’s go ahead and start converting this line integral to a double integral.
So I’ll write my double integral notation here.
I’m going to leave some space over here for our upper and lower limits of integration.
Now partial derivative of q with respect to x is going to be 4x minus partial derivative
of p with respect to y which is 1.
And I’m going to go ahead and put my da in there.
This da is actually going to get replaced because our inner integral is going to represent
y, our outer one is going to represent x.
Rememeber that when that’s the case, because y is the inner integral, we have to have dy
here on the inside and then x is the outer integral so we have to then have dx here on
the outside.
So we replace the da with dy,dx to match inner, inner, outer and outer.
Now all we need is our limits of integration.
This is where this problem gets a little bit trickier than some other Green’s theorem
problems because we’ve got this line integral.
We’ve graphed it here and it represents a triangle.
The line integral bounds a triangular region.
Unfortunately, we can’t just evaluate the area inside the entire triangle because we
essentially have two regions here.
This triangle is cut in half and the reason is because this line here is represented by
a different equation than this line here.
So we need to evaluate this in two parts.
And what we’re going to do is evaluate this region first, half of it, and then we’re
going to evaluate the second region over here.
Our upper limit of integration for this line region is going to be this equation here.
Our lower limit of integration will be the x axis here.
And that’s going to be y equals zero.
For the teal region, I guess you could call it, we’ll have some other function g of
x which we need to find.
That will be our upper limit of integration.
And our lower limit of integration will be this line here which again is going to be
y equals zero, the x axis.
So we just need to find f of x and g of x.
So f of x we know to be the line y equals x.
And the reason is because we have two points on it, (0,0) and (1,1), and we know that that
line intersects or bisects the first and third quadrants exactly.
You could use the equation y equals mx plus b, plug in your points (0,0) and (1,1), and
find that equation y equals x.
And in fact, let’s go ahead and do that for g of x since it’s a little bit more
complicated to find.
So we’ll use this equation y equals mx plus b to find the equation of that line and remember,
m is the slope so we’ll use the two points we have, (1,1) and (2,0).
Remember that we’ll subtract the y coordinates from each other so we’ll say zero minus
1 and then our denominator is going to be 2 minus 1 times x plus b.
When we simplify this, we’ll get -1/1 which is just -1.
So we’ll get negative x plus b.
Then we need to pick either point on the line, plug those in for y and x to solve for b.
So we’ll use (1,1).
We’ll get 1 equals -1 plus b.
Adding 1 to both sides gives us 2 equals b.
So then we see that the equation of this line, g of x, is y equals negative x plus 2.
So we’ll use that y equals negative x plus 2.
Now, we have the equation of f of x and we have the equation of g of x.
That’s all we need to define our limits of integration.
So for the line region, the first thing we’ll do is say that our lower limit of integration
is zero because this part right here, the lower bound is the line y equals zero.
So we’ll put zero there as our lower limit of integration.
The upper limit of integration is defined by this function here, f of x.
That’s the upper limit.
So we already know that that line is y equals x.
So we’\ll go ahead and put x here.
so that’s how we integrate with respect to y.
With respect to x though, realize that we are integrating between the lines x equals
zero and x equals 1.
Therefore, our limits of integration for x are zero and 1.
That, keep in mind, only gives us this part right here, the line region.
We’ve only done half of it.
Now we need to do the second half.
So in order to do that, we have to add to this a second double integral to get this
second half of the region.
So in this case, we’ll use the same function, 4x minus 1.
We’ll have dy and dx.
But in this case, for our limits of integration with respect to y, our lower limit of integration
will be again y equals zero, this line here, the x axis.
So we’ll go ahead and put in zero.
but the upper limit of integration is this line here which is defined by y equals negative
x plus 2.
We found that here.
That’s g of x.
So we’ll say negative x plus 2 is our upper limit of integration.
That’s integrating with respect to y.
Then when we’re integrating with respect to x, remember that now, we’re integrating
between the lines x equals 1 and x equals 2.
So our limits of integration with respect to x are 1 and 2.
So that’s how to set up our integral and that gives us the second half here of this
triangle, the right half of the triangle.
So now we have everything we need and we can just go ahead and evaluate these double integrals.
So to evaluate the double integral, what we’ll do is first evaluate with respect to y.
We’ll evaluate with respect to y because remember, dy is on the inside so we integrate
with respect to y first.
So we’ll go ahead and leave this outer integral because we got to deal with that later.
Remember when we integrate with respect to y, we treat x as a constant and y as the variable.
So our integral becomes 4xy minus y and we’ll be evaluating that on the range zero to x.
We plug zero and x in for y.
We’ll go ahead and leave our dx.
Again, same thing here, we leave our outer integral from 1 to 2.
We evaluate.
We get 4xy minus y and we’ll be evaluating that on the range zero to negative x plus
2, and we go ahead and leave dx there as it is.
So now we can evaluate that definite integral there.
So we’ll leave the zero and the one.
And here for this first integral, remember we’ll be plugging in our upper limit of
integration first, x.
We’re plugging in x for y because we already integrated with respect to y so we’re plugging
in x for y.
When we do that, we’ll get 4x squared minus x minus then we plug in our lower limit of
integration and of course we’ll get 4x times zero minus zero and we go ahead and leave
our dx.
Then we add to that our second integral so we got 1 to 2.
Now we’ll be plugging in this upper limit of integration, negative x plus 2.
So we’ll get 4x times negative x plus 2 minus negative x plus 2.
And if we have plugged in zero, of course we get minus zero minus zero so that’s just
going to give us zero, then dx.
So we need to simplify before we move on to our next step.
We’ll have 4x squared minus x.
This is going to become zero.
So we just have dx.
Then we have plus from 1 to 2, that’s going to give us a negative 4x squared right here.
Then we have plus 8x plus another x is going to be plus 9x and then we’ll have a minus
2 dx.
Alright.
So now that we’ve plugged in those limits of integration with respect to y, we can integrate
again each of these with respect to x and evaluate the definite integral the second
time.
So here, when we integrate with respect to x, we’ll get 4/3 x cubed minus 1/2 x squared
and we’ll be evaluating that on the range zero to 1.
And when we add to that, we get negative 4/3 x cubed plus 9/2 x squared minus 2x and we’ll
be evaluating that on the range 1 to 2.
Okay, so now, we’ll go ahead and plug in our upper limits of integration, 1, first
then our lower limit of integration, zero.
So 4/3 times 1 cubed is just going to give us 4/3 minus 1/2 times 1 squared gives us
just 1/2.
When we plug in our lower limit of integration, zero, this will give us zero and zero so we
can just go ahead and leave that as it is.
Then, we’ll add and plug in our upper limit of integration 2, so 2 cubed here gives us
8 times a negative 4/3 gives us negative 32/3.
2 squared here gives us 4 times 9/2 gives us 36/2.
Then 2 times the negative 2 gives us a minus 4.
So that our upper limit of integration.
And now we need to plug in our lower limit of integration so we’ll subtract and we’ll
get negative 4/3 times 1 cubed which is just 1 obviously, plus 9/2 times 1 squared is still
1 minus 2 times 1 is just a negative 2.
Okay.
So now that we’ve evaluated the entire definite integral, we can go ahead and simplify and
I’m going to go ahead and put everything over 6 because we’ve got some things over
3 and some things over 2.
So I’m going to find a common denominator of 6 and multiply here as I go.
So 4/3 becomes 8/6 minus 1/2 becomes 3/6.
Then we’ll get a minus here, 64/6, plus multiply that by 3 and we’ll get 108/6 minus
24/6 is what we get for 4.
Then because we have a negative here and a negative here, we’ll get a positive 8/6.
And negative times 3 gives us 27/6.
And then a positive 2 is going to be plus 12/6.
And if we evaluate that on our calculator, what we’ll get is 18/6 which of course is
going to be equal to 3.
And so our final answer is 3.
And that should make sense to us.
If we look at the vertices of our triangle, (0,0), (1,1), and (2,0), we’ve got essentially
in this first triangle over here 1 and 1/2 units of area and in the second triangle over
here, another 1 and 1/2 units of area.
So our area is 3.
The area defined by the line integral that defines this triangle is 3.
So that’s it.
That’s our final answer.
I hope that video helped you guys and I will see you in the next one.
Bye!