Question

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?

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

i think you are confusing a few things

Answer 5

\[\int\int \frac{x^2+y^2}{xy} dy.dx\]

\[\int\int \frac{x^2}{xy}+\frac{y^2}{xy} dy.dx\]

\[\int\int \frac{x}{y}+\frac{y}{x} dy.dx\]

\[\int\int xln(y)+\frac{y^2}{2x} \ dx\]etc...

if that equates to your idea, then try it out

Answer 6

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

Answer 7

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

Answer 8

oh, i wrote it for the practice:)

Answer 9

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

Answer 10

:) heh

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

no

Answer 17

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

Answer 18

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

Answer 19

Don't worry.

Answer 20

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

Answer 21

thta's some crazy cartesian-polar hybrid

Answer 22

hahaha

Answer 23

but its good to see you thinking outside the mathical box

Answer 24

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

Answer 25

thank you all for the help!

Answer 26

welcome :)