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## AP®︎/College Calculus BC

### Course: AP®︎/College Calculus BC > Unit 8

Lesson 5: Finding the area between curves expressed as functions of y# Horizontal area between curves

We can use a definite integral in terms of 𝘺 to find the horizontal area between curves of two functions of 𝘺.

## Want to join the conversation?

- At2:12, why do we know what function hold a bigger value than the other one? Why we let f(y) - g(y)?(13 votes)
- Well, in case you try to visualize this by turning the graph by 90 degrees keeping the positive x-axis side up, you would see why. Or, another way to think about this is that since both functions have x as their dependent variable, so what function gives out a larger x value for the same y input, and when both the x values that you get from each of the functions from the same y value coincides, you can be pretty confident about the fact that the y values must be the upper and lower bounds of integration. Of course, by checking which intersection point has a larger y.

Cheers!(14 votes)

- i cant know why one is the upper or lower curve in this one is??(5 votes)
- It's just about recognizing which function takes on the higher x-value. As we learned about in Vertical Areas between Curves, the function with the higher
*y-value*is the upper curve. Here, since we are taking the horizontal areas between curves, we have to think about the**x-value**and we get whichever one is the upper and lower functions.(11 votes)

- Is it because you subtract g(y) from f(y) that you don't need special treatment of the area that falls below the x-axis and would otherwise generate a negative number?(1 vote)
- Remember that we are integrating over 𝑦 and not over 𝑥, which means that we're looking at the area between the curve and the 𝑦-axis.

We have "positive area" to the right of the 𝑦-axis, and "negative area" to the left. However, since 𝑓(𝑦) ≥ 𝑔(𝑦) over the interval 𝑦 ∈ [−2, 3] and we're subtracting 𝑔(𝑦) from 𝑓(𝑦) the result is always positive, and we'll end up with the area between the two curves.(5 votes)

- just wanted to know why sal equated both the equations at "2:45",

I wonder why we didn't take the derivative of both the equations and equate it with zero as the endpoints of those contain a vertical slope(2 votes)- Because the limits of integration are where the graphs INTERSECT, not the endpoints. The endpoints are not the same as the intersections. If we were integrating 2 y= graphs, you'd be correct. Try looking at the graph sideways; that might help.(2 votes)

- in this question, can you take the inverse function and write it in terms of y and then solve for it?(2 votes)
- I think that is a valid method.(1 vote)

- I think the final result should be 41(0 votes)

## Video transcript

- [Instructor] So I have
two curves graphed here. And we're used to seeing things
where y is a function of x, but here we have x as a function of y. In fact, we can write this top expression as being a function of y. And this second one, just
to make it different, we could view this as g of y. Once again, it is a function of y. And what we're concerning
ourselves with in this video is how do we find this area in this light blue color
between these two curves? And I encourage you to pause the video and try to work through it. All right, so a huge hint here is we're going to want to
integrate with respect to y, a definite integral where
our bounds are in terms of y. So for example, this is, this lower point of
intersection right over here, this would be our lower
bound in terms of y. Let's call that y one. And then, this up here, this would
be our upper y bound. So if we think about where do
these two curves intersect, and we look at the y-coordinates
of those intersections, well, that gives us two nice
bounds for our integral. So we're gonna take our
integral from y one to y two, from y one to y two, y two. We're going to integrate
with respect to y, dy. And so what are we going to sum up? Well, when we integrate,
we can think about taking the sum of
infinitely thin rectangles. And in this case, it would
infinitely flat rectangles since we're thinking about dy. So dy would be the height
of each of these rectangles. And what would be, in this case, the width or the length of this
rectangle right over here? Well, over this interval,
from y one to y two, our blue function, f of y, takes on larger x-values than g of y. So this length right over
here, this would be f of y, f of y, this x-value minus this x-value, minus g of y. So this is going to be f of y minus g of y, g of y. Well, we know what f of y and g of y are. Really, the trickiest part is, is figuring out these
points of intersection. So let's think about where
these two curves intersect. They are both equal to x, so we can set these two y
expressions equal to each other. So we know that negative, let me do it in that other color, so we know that negative
y squared plus three y plus 11 is going to be equal to this, is going to be equal to y squared plus y minus one. So let's just subtract all
of this from both sides so that on the right side we have a zero and on the left side we
just have a quadratic. So let's subtract y squared. Let's subtract y. And then subtract negative
one, which is just adding one. And over here, we're going
to do the same thing, minus y plus one. And what we are left
with is going to be a, hopefully, a straightforward quadratic. So let's see, this is going
to be negative two y squared plus two y. Am I doing that right? Yep, plus two y, plus 12 is equal to zero. And then this over here I can
factor out a negative two, and I get negative two
times y squared minus y minus six is equal to zero. This we can factor from inspection. What two numbers, when we
add, equal negative one? When we take their product,
we get negative six? Well, that would be
negative three and two. So this is going to be negative
two times y minus three times y plus two. That's just straightforward
factoring a polynomial, a quadratic. Did I do that right? Yep, that looks right, is equal to zero. So what are the points of intersection? The points of intersection are going to be y is equal to three and y
is equal to negative two. So this right over here is
y is equal to negative two, and then the upper bound
is y is equal to three. So now we just have to evaluate
this from negative two, all the way until three. So let's do that. I'm gonna clear this out, so I get a little bit of real estate. So this is equal to the integral from negative two to three of negative y squared, plus three y, plus 11, minus all of this stuff. So if we just distribute
a negative sign here, it's minus y squared, minus y, plus one, and then we have a dy, dy. This is equal to the
definite integral from negative two to positive three of, let's see, negative y
squared minus y squared, negative two y squared, and then three y minus y
is going to be plus two y. And then 11 plus one, plus
12, we saw this just now, when we were trying to solve for y, dy. And so what is that going to be equal to? Well, we just take the
antiderivative here. This is going to be,
let's see, negative two. Let's increment the
exponent, y to the third. Divide by that exponent,
reverse power rule, plus two y squared divided by
two, which is just y squared, just the reverse power rule, and then plus 12 y. And we're going to evaluate that at three and at negative two. So if we evaluate that at
three, we are going to get, let's see, negative two times 27 over three, plus nine, plus 36. And then we are going to
want to subtract, minus all of this evaluated at negative two. So it's going to be negative
two times negative eight over three, plus four, minus 24. So we just have a little bit
of mathematics ahead of us. So let's see, this is going to be 27
divided by three is nine. So this is negative 18. Negative 18 plus nine is going to be negative nine, plus 36, all of that is
going to be equal to, so the stuff in blue is equal to 27. Right, did I do that right? Get negative 18 plus nine, yep, 27. And then all the stuff in red over here, we have, this is going to be
negative times a negative, so it's 16 over three,
plus four, minus 24. So that is going to be 16 over three, and then minus, minus 20, minus 20. But then we have this
negative out here, so it's, if we distribute that, we'll get plus 20, minus, we could say
instead of 16 over three, we could rewrite that as 5 1/3, minus 5 1/3. And so what is that going to get us? Let me scroll down a little
bit, or let me go to the right, so I have a little bit
more real estate here. So then that is going to be equal to, get a minor drum roll here, 47 minus five minus 1/3, which is equal to, let's see,
47 minus five, that's 42, minus 1/3 is, now we get a drum roll, 41 2/3. And we are done.