double integral
\[x^{3}y \] for R bounded by y = x^2 and y = 2x please help
@missylulu
firstly trace these two curves and find the point of intersection and check the variation for x and y
x = 0, 2!
ya x=0,2 and corresponding to that y=0,4 so the two points are (0,0) and (2,4) right?
you draw the graph...........trace these two curves...
you dont use y=0,4 you need to integrate from x^2 to 2x or vice versa you need to integrate the y first because y is dependant on x
i agree that you should start by drawing a graph to determine the order in which you integrate the boundaries
do you see which one is lower? like are the limits bottom to top?? fr om the graph
\[\int\limits_{0}^{2}\int\limits_{2x}^{x^2} {x^3y} dy dx\]
|dw:1358843349591:dw|
oops i wrote the integral incorrectly
x is from 0 to 2 so we only look at that section when considering whether x^2 or 2x should be on top of the integral sorry my termnology may be horrible and i hope you can somewhat understand what im saying anyways when you actually draw the graph, the area that you are integrating is between 2x and x^2 2x being on top
ok ok! so like for integrals where it gives you some equation you just graph it to see what to integrate it by? you dont have to create new limits or something?
like that interchanging limits or something lol sorry
well, this problem is 3 dimensional z=x^3y but when determining the order for integrating dependant limits just graph the boundaries to determine order in which you're integrating
on a side note, using your problem as an example, if you were integrating from 0,4 or any other number greater than 2 you would have to break it up into 2 integrals because past x=2, x^2 would be above 2x so basically integrating from 0 to 2, then 2 to 4
ohhh okok! btw youre not a completeidiot i am LOL THANK YOU ^^
which is why drawing a graph is highly suggested, unless you can picture it in your head but even then you're bound to screw up from time to time
no problem
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