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### Course: AP®︎/College Calculus AB>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?

• At , why do we know what function hold a bigger value than the other one? Why we let f(y) - g(y)?
• 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!
• i cant know why one is the upper or lower curve in this one is??
• 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.
• 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.
• just wanted to know why sal equated both the equations at "",
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
• 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.
• in this question, can you take the inverse function and write it in terms of y and then solve for it?
• I think that is a valid method.
(1 vote)
• If your allowed to use calculator on your test a trick is to plug the function for example if the question gives x=-y^2+3y+11 as y=-x^2+3x+11 and the other equation x=y^2+y-1 as y=x^2+x-1 and use the calculator to find the x-values of the point of intersection. the x values of this are the same as the y values of intersection in the original equations. You can just plug this in to your equation and you will get the same answer.
(1 vote)
• why do we put f(y) in front, why is it not g(y)-f(y)?
(1 vote)
• I think the final result should be 41