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anonymous
 4 years ago
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?
anonymous
 4 years ago
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?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0using the cos^2x + sin^2y identity

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

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

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

amistre64
 4 years ago
Best 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ah 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

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0just 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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You can use that identity to simplify integrals but you have to be carefull.

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[x^2+y^2\neq \cos^2x+\sin^2x\]

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

TuringTest
 4 years ago
Best 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0er 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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0you 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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.0thta's some crazy cartesianpolar hybrid

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thank you all for the help!
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