Thread: Find the Area between the curves

Hi, I have two calculus problems that I need help on but do not know how to solve. Any help would be appreciated!

Find the area enclosed by the curves:

(a). y = x^2 - 1, y = -1, x = 1, x = 2

(b). x = y^4, y = sqrt(2-x), y = 0

Here is a picture in case I was not clear enough:

http://i44.tinypic.com/2zq7yia.png

Now I know that this has to deal with the usage of integrals to figure out the area, however, I do not know exactly how to go about it, or feel that I am doing so incorrectly. I have uploaded a pdf of the work I have done thus far, getting the answer 7/3 for the first question. Is this wrong or am I doing the work incorrectly?

a)





b) is a little trickier. I haven't seen your document; however, it is always a good idea to graph the curves and identify the region you are trying to calculate first. For b) you can split things up and calculate two integrals one from 0 to 1 and another from 1 to 2 using the appropriate curves, and then add them together for the area of your region. See how you go.

The only "hard" part of (b) is determining where those two curves cross. If we square both sides we get or . That could be very difficult to solve in general but fortunately, it is easy to see that x= 1 is a solution.

The area below those two graphs and above y= 0 is given by


The first of those should be very easy to integrate. To integrate the second, let y= 2- x so that dy= -dx, when x= 1, y= 2- 1= 1, when x= 2, y= 2- 2= 0, and you have

The first curve of b) is not though. You need to change it to to y = before you integrate from 0 to 1 to create the same region, otherwise you can change them to and and integrate between 0 and 1 subtracting the bottom curve from the top.

Okay thank you all so much for the help! So Terror as long as I make y=x^4 integrable to y=2-x^2 I can then use what Halls suggested? Otherwise I would have to integrate from y=x^(1/4)? Which way would you think would be easier?

Then from that I also do what Halls suggested in the second portion of his explanation and integrate those values and then subtract from the values of the first step?

You wouldn't use that method if you were going to use the x=y^4 function as it is.

The best way to understand what is going on is just to plot the graphs in some software or use wolfram alpha. It is easier to see the different ways you might attack the problem that way.

What I did the second time around was change the second equation so they were both in terms of the same variable:

you have and

to get them both in terms of , take



square both sides





Setting the new equation = to , you will find that they intersect at , but you only want the region from y=0 to y=1. Then you take the integral as per normal subtracting the bottom curve from the top curve dy.

Whether you use y=x^4 and y=2-x^2 between x=0 and x=1, or x=y^4 and x=2-y^2 between y=0 and y=1 wont matter, the size of the region is the same.



The other method we were using calculates the region along a perpendicular axis, so no curve is really above or below the other one, in which case its easiest to calculate the areas separately. The two regions press up against one another to form the entire region you are calculating.

Good luck.

Okay thank you for the clarification. So I went and integrated what you have given me and came up with this answer. Is this the area of the curve then? Or is there still more work to be done?

Yah, that is the area between the two curves from y=0 to y=1. You can just do it all as one integral if you like; no need to split them up - that's up to you though.