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ksmith197
Group Title
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?
 2 years ago
 2 years ago
ksmith197 Group Title
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?
 2 years ago
 2 years ago

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ksmith197 Group TitleBest ResponseYou've already chosen the best response.0
using the cos^2x + sin^2y identity
 2 years ago

ksmith197 Group TitleBest ResponseYou've already chosen the best response.0
\[\int\limits_{1}^{4}\int\limits_{1}^{2}(x^2+y^2)/xy)dydx\]
 2 years ago

ksmith197 Group TitleBest ResponseYou've already chosen the best response.0
why can't the x^2 + y^2 be set = 1
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
i think you are confusing a few things
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
\[\int\int \frac{x^2+y^2}{xy} dy.dx\] \[\int\int \frac{x^2}{xy}+\frac{y^2}{xy} dy.dx\] \[\int\int \frac{x}{y}+\frac{y}{x} dy.dx\] \[\int\int xln(y)+\frac{y^2}{2x} \ dx\]etc... if that equates to your idea, then try it out
 2 years ago

ksmith197 Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

ksmith197 Group TitleBest ResponseYou've already chosen the best response.0
was just kinda wondering when you can or can't use the x^2+y^2 = cos[x]^2 + sin[y]^2 = 1 identities
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
oh, i wrote it for the practice:)
 2 years ago

ksmith197 Group TitleBest ResponseYou've already chosen the best response.0
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 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
(..)^2 is not generally equativalent to the function trig^2(...)
 2 years ago

Nodata Group TitleBest ResponseYou've already chosen the best response.3
You can use that identity to simplify integrals but you have to be carefull.
 2 years ago

ksmith197 Group TitleBest ResponseYou've already chosen the best response.0
yeah, the integral didn't come out right in mathematica. Or, could you use it but you have to change the limits of integration?
 2 years ago

Nodata Group TitleBest ResponseYou've already chosen the best response.3
\[x^2+y^2\neq \cos^2x+\sin^2x\]
 2 years ago

ksmith197 Group TitleBest ResponseYou've already chosen the best response.0
ah right but x^2/r^2 + y^2/r^2 = cos[x]^2 + sin[y]^2 = r^2 right?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
\[x=r\cos\theta\]\[y=r\sin\theta\]\[x^2+y^2=r^2\cos^2\theta+r^2\sin^2\theta=r^2\]I'm not sure what you are trying to get out of that
 2 years ago

ksmith197 Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

Nodata Group TitleBest ResponseYou've already chosen the best response.3
Don't worry.
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
thta's some crazy cartesianpolar hybrid
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
but its good to see you thinking outside the mathical box
 2 years ago

ksmith197 Group TitleBest ResponseYou've already chosen the best response.0
ah i haven't read the chapter on polar stuff yet, so maybe i'll remedy things in the coming week :)
 2 years ago

ksmith197 Group TitleBest ResponseYou've already chosen the best response.0
thank you all for the help!
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
welcome :)
 2 years ago
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