Integral area problem

[SockCymbal]

Ok, the instructions are to find the area of the region enclosed by the curves...

 

x=y^3-4y^2 +3y and x=y^2-y

 

the coordinates aren't given, but I have worked the y coordinated out to be 0, 1, and 3. This problem throws me off because of its in x in terms of y, instead of vice versa. Can anyone help me out as to how to approach graphing it. Thanks

[Tom Mattson]

Just interchange y and x. What this amounts to is a rotation of the figure, which does not change its area.

[Dave]

The best advice I can give you is to draw it out and observe what integrals you need to consider. A plot is always useful in these cases.

[Tom Mattson]

He seems to know that he has to draw it out. What I think he's saying is that he doesn't know how to, because for both curves x is given as a function of y, and he learned his curve sketching techniques when the opposite was the case. That's why I say to interchange coordinates. He should then be able to easily plot the figure, and it will have the correct area.

[Pat Says]

Might switching the variables produce a sign error though?

 

Edit: arg... Im too tired... it won't matter with area.

[Tom Mattson]

You are onto something though. I made a tiny mistake in my first post. Switching x and y is not just a rotation. It is a rotation and a reflection.

 

So say you have both curves plotted on coordinate system with the axes in "standard" position. If you rotate counterclockwise by 90 degrees, then reflect through the new x-axis (which is now vertical), then you get the same figure you would have gotten if you had just switched x and y in the first place. And the area is still the same, of course.

[Pat Says]

Yeah, otherwise you would have to switch witch integral is being subtracted from the other ( I think..)

[SockCymbal]

i tried switching the variables, but that also would change the parameters of the integral...i believe in this case though the integral would be from -1 to 0. could someone double check this for me though?

[Tom Mattson]

The limits of integration will not change. Why would they? Instead of setting y^3-4y^2 +3y equal to y^2-y and solving, you are now setting x^3-4x^2 +3x equal to x^2-x and solving. It should be easy to see that you will get the same solutions.

[SockCymbal]

i thought they would change because one you rotate the graph, the area enclosed by the graphs would also rotate to a new set of x values...

[Tom Mattson]

You do get a new set of x-coordinates for the points of intersection after the transformation. They are precisely the old y-coordinates, which you already found.