Integration: Finding Area Between Curve along X and Y axis Questions

  • Thread starter Ahlahn
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Hello Everyone,
Below are three questions relating to the integration between curves that have been bugging me for some time now. I know how to integrate between curves. I know how to integrate between x and y, but when given equations like the ones below I am lost as to how to figure out 2 things

1. What the graphs look like(so I know which one is left/right, top/bottom)
2. The points of intersection(to figure out the interval [a,b])

You don’t have to answer all of my questions(Though that would be Awesome), but it would really help if you could point me in the right direction by telling me what I’m doing wrong so I won’t keep running into the same problems again. Thanks so much!

Question 1

Find the area of the region enclosed by the semicubical parabola y2 = x3 and the line x = 3
.

I am having a lot of trouble with questions like this one. I am assuming that the question is asking me to integrate along the y axis between the curves x=y^2/3 and x=3, but what is the interval??? How can I find the interval when given functions like these? What am I doing wrong?

Question 2

Sketch the region enclosed by the curves and compute its area as an integral along the x- or y-axis.
x = y3 - 18y, y + 7x = 0


I rewrote it as ...X = y^3 – 18y and x = -(1/7)y
Like the previous question, I have trouble figuring out how the graphs look like, which one lies on top or side- so I am clueless as to whether I should integrate along the x or y…

Question 3


Find the area enclosed by the curves listed below as a function of c.
y = c - x^2
y = x^2 - c


Okay so I plugged a random constant for c and found that the function y = c-x^2 lies on top, therefore we are integrating along the x axis. I then tried looking for the points of intersection, and failed. How can I find the interval if c is unknown?! O_O

Again, thanks everyone in advance!
 

Answers and Replies

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1. Unfortunately, knowing what graphs look like is something that sort of just comes with experience- you need to know some of the basic ones like x^2, x^3, x^(1/2), 1/x, etc. and how variations on these affect the graph, like how you can graph (x-3)^2 by knowing what x^2 looks like. I learned these things in pre-calc. If you didn't learn these things, I would suggest that either you teach yourself or rely on your graphing calculator.

2. Figuring out the area you're dealing with, and the points of intersection, is easiest if you have a rough idea of what the graphs of your function look like. It's going to be harder for you if you don't know this.
Generally when a problem gives two curves, you can assume that they intersect at two points. Find these points of intersection by setting the functions equal to each other. To see which is above, pick any point in between the points of intersection and see which is greater. This method does not necessarily apply to area problems that specify more than two functions, though. You might have something that says "Find the area between f(x), g(x), x=a, x=b", in which case your limits of integration are a and b.

Question 1: Since this problem gives you two curves, you can first assume that they intersect at two points. So set them equal, [tex]y^{2/3}=3[/tex] and you will find your interval (taking into account that a square root will give you a positive and negative value). Alternatively, if you recognize that [tex]x=y^{2/3}[/tex] is symmetric about the x-axis, you can integrate from 0 to the positive square root value and multiply the integral by 2.
Also, you don't necessarily have to integrate along the y-axis. Again, if you recognize that it's symmetric about the x-axis, since [tex]y=0^{3/2}=0[/tex] you can integrate with respect to x from 0 to 3 and multiply the integral by 2.
I uploaded a picture to illustrate what I mean.
 

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You also need to realize that if you integrate with respect to y in Question 1, your integral is going to be
[tex]\int\left(3 - y^{2/3}\right)dy[/tex]
which is slightly more complicated than if you integrated with respect to x.
 

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