SockCymbal Posted June 14, 2005 Share Posted June 14, 2005 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 Link to comment Share on other sites More sharing options...
Tom Mattson Posted June 14, 2005 Share Posted June 14, 2005 Just interchange y and x. What this amounts to is a rotation of the figure, which does not change its area. Link to comment Share on other sites More sharing options...
Dave Posted June 14, 2005 Share Posted June 14, 2005 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. Link to comment Share on other sites More sharing options...
Tom Mattson Posted June 14, 2005 Share Posted June 14, 2005 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. Link to comment Share on other sites More sharing options...
Pat Says Posted June 15, 2005 Share Posted June 15, 2005 Might switching the variables produce a sign error though? Edit: arg... Im too tired... it won't matter with area. Link to comment Share on other sites More sharing options...
Tom Mattson Posted June 15, 2005 Share Posted June 15, 2005 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. Link to comment Share on other sites More sharing options...
Pat Says Posted June 15, 2005 Share Posted June 15, 2005 Yeah, otherwise you would have to switch witch integral is being subtracted from the other ( I think..) Link to comment Share on other sites More sharing options...
SockCymbal Posted June 15, 2005 Author Share Posted June 15, 2005 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? Link to comment Share on other sites More sharing options...
Tom Mattson Posted June 15, 2005 Share Posted June 15, 2005 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. Link to comment Share on other sites More sharing options...
SockCymbal Posted June 15, 2005 Author Share Posted June 15, 2005 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... Link to comment Share on other sites More sharing options...
Tom Mattson Posted June 15, 2005 Share Posted June 15, 2005 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. Link to comment Share on other sites More sharing options...
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