anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
using the cos^2x + sin^2y identity
anonymous
  • anonymous
\[\int\limits_{1}^{4}\int\limits_{1}^{2}(x^2+y^2)/xy)dydx\]
anonymous
  • anonymous
why can't the x^2 + y^2 be set = 1

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amistre64
  • amistre64
i think you are confusing a few things
amistre64
  • 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
  • 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
  • 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
  • amistre64
oh, i wrote it for the practice:)
anonymous
  • 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
  • anonymous
:) heh
amistre64
  • amistre64
(..)^2 is not generally equativalent to the function trig^2(...)
anonymous
  • anonymous
You can use that identity to simplify integrals but you have to be carefull.
anonymous
  • 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
  • anonymous
\[x^2+y^2\neq \cos^2x+\sin^2x\]
anonymous
  • anonymous
ah right but x^2/r^2 + y^2/r^2 = cos[x]^2 + sin[y]^2 = r^2 right?
TuringTest
  • TuringTest
no
TuringTest
  • 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
  • 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
  • anonymous
Don't worry.
TuringTest
  • 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
  • TuringTest
thta's some crazy cartesian-polar hybrid
anonymous
  • anonymous
hahaha
amistre64
  • amistre64
but its good to see you thinking outside the mathical box
anonymous
  • anonymous
ah i haven't read the chapter on polar stuff yet, so maybe i'll remedy things in the coming week :)
anonymous
  • anonymous
thank you all for the help!
TuringTest
  • TuringTest
welcome :)

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