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

- chestercat

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- anonymous

using the cos^2x + sin^2y identity

- anonymous

\[\int\limits_{1}^{4}\int\limits_{1}^{2}(x^2+y^2)/xy)dydx\]

- anonymous

why can't the x^2 + y^2 be set = 1

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## More answers

- amistre64

i think you are confusing a few things

- amistre64

\[\int\int \frac{x^2+y^2}{xy} dy.dx\]
\[\int\int \frac{x^2}{xy}+\frac{y^2}{xy} dy.dx\]
\[\int\int \frac{x}{y}+\frac{y}{x} dy.dx\]
\[\int\int xln(y)+\frac{y^2}{2x} \ dx\]etc...
if that equates to your idea, then try it out

- anonymous

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

- anonymous

was just kinda wondering when you can or can't use the x^2+y^2 = cos[x]^2 + sin[y]^2 = 1 identities

- amistre64

oh, i wrote it for the practice:)

- anonymous

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

- anonymous

:) heh

- amistre64

(..)^2 is not generally equativalent to the function trig^2(...)

- anonymous

You can use that identity to simplify integrals but you have to be carefull.

- anonymous

yeah, the integral didn't come out right in mathematica. Or, could you use it but you have to change the limits of integration?

- anonymous

\[x^2+y^2\neq \cos^2x+\sin^2x\]

- anonymous

ah right but x^2/r^2 + y^2/r^2 = cos[x]^2 + sin[y]^2 = r^2 right?

- TuringTest

no

- TuringTest

\[x=r\cos\theta\]\[y=r\sin\theta\]\[x^2+y^2=r^2\cos^2\theta+r^2\sin^2\theta=r^2\]I'm not sure what you are trying to get out of that

- anonymous

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

- anonymous

Don't worry.

- TuringTest

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

- TuringTest

thta's some crazy cartesian-polar hybrid

- anonymous

hahaha

- amistre64

but its good to see you thinking outside the mathical box

- anonymous

ah i haven't read the chapter on polar stuff yet, so maybe i'll remedy things in the coming week :)

- anonymous

thank you all for the help!

- TuringTest

welcome :)

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