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ksmith197
Easy calculus question: if you had something like say a double integral integral[integral (x^2+y^2)/(xy) dy]dx why can't you set x^2 + y^2 = 1?
using the cos^2x + sin^2y identity
\[\int\limits_{1}^{4}\int\limits_{1}^{2}(x^2+y^2)/xy)dydx\]
why can't the x^2 + y^2 be set = 1
i think you are confusing a few things
\[\int\int \frac{x^2+y^2}{xy} dy.dx\] \[\int\int \frac{x^2}{xy}+\frac{y^2}{xy} dy.dx\] \[\int\int \frac{x}{y}+\frac{y}{x} dy.dx\] \[\int\int xln(y)+\frac{y^2}{2x} \ dx\]etc... if that equates to your idea, then try it out
ah sorry didn't mean to make you write all of that out that's actually what the original equation was, i just moved evertying around to try something out
was just kinda wondering when you can or can't use the x^2+y^2 = cos[x]^2 + sin[y]^2 = 1 identities
oh, i wrote it for the practice:)
just figured that since we are using double integrals, they are equations with x and y, so i was thinking you could use identities to maybe simplify things
(..)^2 is not generally equativalent to the function trig^2(...)
You can use that identity to simplify integrals but you have to be carefull.
yeah, the integral didn't come out right in mathematica. Or, could you use it but you have to change the limits of integration?
\[x^2+y^2\neq \cos^2x+\sin^2x\]
ah right but x^2/r^2 + y^2/r^2 = cos[x]^2 + sin[y]^2 = r^2 right?
\[x=r\cos\theta\]\[y=r\sin\theta\]\[x^2+y^2=r^2\cos^2\theta+r^2\sin^2\theta=r^2\]I'm not sure what you are trying to get out of that
er yeah, sorry messed that up. it would still work out though right?? because x^2/1 + y^2/1 = (1)^2 cos^2 + sin^2 = 1^2 blehhh i think i'm mixing this all up
you are thinking something like\[x^2+y^2=r^2\implies{x^2+y^2\over r^2}=1\]perhaps? That would totally screw up your coordinate system though, to put that in the integral
thta's some crazy cartesian-polar hybrid
but its good to see you thinking outside the mathical box
ah i haven't read the chapter on polar stuff yet, so maybe i'll remedy things in the coming week :)
thank you all for the help!