The region D above lies between the graphs
of y equals two minus the quantity x plus three squared
and y equals negative two plus 1/9
times the quantity x plus five cubed.
We're going to be describing the region in two ways,
first using a top and bottom boundary,
then a right and left boundary.
Before we do this, though,
let's match up these functions to the graphs.
We should be able to recognize this first function:
y equals two minus the quantity x plus three squared
would be a quadratic function.
And if we expanded this,
the coefficient of the x squared term would be negative,
and therefore the graph is a parabola opening down,
which means this piece of the graph
would have to come from the function, y equals two
minus the quantity x plus three squared.
Which means the bottom graph
would have to come from the function given here,
y equals negative two
plus 1/9 times the quantity
x plus five raised to the third.
Let's also identify some of the points.
Notice this point here would have coordinates
negative five comma negative two.
This point here would have coordinates
negative two comma one.
And notice the vertex of the parabola here
would have coordinates negative three comma two.
Now take a look at number one.
If we visualize the region having top and bottom boundaries,
express the boundaries as functions of x
and provide the interval of x values
that cover the entire region.
So notice how the top boundary would be
the graph of a parabola here.
And we already know it's y equals two minus
the quantity x plus three squared,
which is already expressed as a function of x,
which means g sub two of x, the top boundary,
would be equal to two minus the quantity
x plus three squared.
And the bottom boundary is this graph here,
which we already know is y equals negative two
plus 1/9 times the quantity x plus five cubed,
which is already a function of x.
Which means g sub one of x would be negative two
plus 1/9 times the quantity
x plus five raised to the third.
Now we're asked to determine
the interval of x values that covers this region.
And that's why it's important to have these key points here.
Notice how the leftmost point of the region
has an x value of negative five.
The rightmost point has an x value of negative two.
Which means the x values that cover this region
would be the closed interval
from negative five to negative two.
For number two, if we visualize the region
having right and left boundaries,
then the right boundary must be defined piece-wise.
Express each as a function of y for the provided integrals
of y values that cover the entire region.
So notice how for the right boundaries,
we have the closed interval from one to two for y
and then from negative two to positive one.
Let's take a look at why this is on the next screen.
If we analyze this graph closely,
notice how the right boundary, starting at the top,
would be from here to here
along the quadratic function
and then from here to here
along the cubic function.
So here's the closed interval for y
from positive one to positive two.
And here's the closed interval for y
from negative two to positive one.
And we need to express these as functions of y,
which means we need to solve both of these equations for x.
So let's begin by doing this.
If we have y equals two
minus the quantity x plus three squared,
to solve this for x let's go ahead
and add this quantity to both sides
and also subtract y from both sides.
So we'd have the quantity x plus three squared,
would be equal to two minus y.
And now we're going to take the square root
of both sides of the equation.
But when we do this,
remember we're going to have plus or minus
the quantity two minus y on the right.
So now we have the quantity x plus three
equals plus or minus the square root of two minus y.
And now we'll go ahead and subtract three on both sides.
So we have x equals plus or minus
the square root of the quantity two minus y, minus three.
So notice how we could have x sub one equals
positive square root the quantity two minus y, minus three.
We could also have x sub two
equals negative the square root of the quantity two minus y,
minus three.
So we need to decide which of these two functions of y
give us the left side or right side of the parabola,
because we only want the right side.
To do this, we'll evaluate both equations
at, let's say, y equals one.
So if y is equal to one using this equation,
we would have the square root of two minus one,
or the square root of one, is one,
minus three, which is negative two for x.
But if we substitute one for y here,
we're going to have the opposite of the square root of one,
which is negative one,
minus three, which is negative four.
Which means the equation x sub one
would give us this point here
on the right side of the parabola.
And notice that negative four comma one is this point here
on the left side of the parabola.
When we're defining the right boundary,
we want the right side of this parabola.
So we're going to be using this equation here
when y is on the closed interval from one to two.
And again, if x is equal to the square root
of the quantity two minus y,
minus three,
because x is equal to f of y,
we can say f sub two of y
is equal to the square root of two minus y,
minus three.
So going back to the previous slide,
again this is going to be
the square root of two minus y,
minus three.
And now for the right boundary
when y is on the closed interval from negative two to one.
Again, this is on the cubic function here.
We need to solve this cubic function for x
to have a function of y.
So now, starting with y equals negative two
plus 1/9 times the quantity x plus five cubed,
solving for x, let's first add two to both sides
so that y plus two equals
1/9 times the quantity x plus five cubed.
Now multiply both sides by nine.
So we'd have nine times the quantity y plus two
equals the quantity x plus five to the third.
To undo the cube, we'd take the cube root of both sides.
We could also raise both sides to the 1/3 power.
So now we have the cube root
of nine times the quantity y plus two
equals just x plus five.
When taking the cube root of both sides,
remember we do not need a plus or minus on the left.
So the last step, we subtract five on both sides.
And again, x is equal to f of y.
So f sub two of y over this interval
is equal to the cube root of nine
times the quantity y plus two minus five.
And our last step is to define the left boundary
on the closed interval from negative two to positive.
And the last step is to define the left boundary
when y is on the closed interval from negative two to two.
So notice how the left boundary
is just along this one function here,
which, remember, is the parabola.
But now it's the left side of the parabola,
which, remember, would be given
by this equation here, x sub two.
'Cause remember when y was one,
we determined that the x coordinate
would have to be negative four on the left side.
So, using this equation now,
the left boundary is going to be f sub one of y
equals negative square root of the quantity two minus y,
minus three.
So, going back to our question,
again f sub one of y is equal to
negative the square root of the quantity two minus y,
minus three.
So we can see in this question,
defining the region using left and right boundaries
is much more involved than using top and bottom boundaries.
And this becomes important when we begin to set up
double integrals to determine the limits of integration.
I hope you found this helpful.