## 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?

1. anonymous

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

2. anonymous

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

3. anonymous

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

4. amistre64

i think you are confusing a few things

5. 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

6. 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

7. 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

8. amistre64

oh, i wrote it for the practice:)

9. 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

10. anonymous

:) heh

11. amistre64

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

12. anonymous

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

13. 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?

14. anonymous

$x^2+y^2\neq \cos^2x+\sin^2x$

15. anonymous

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

16. TuringTest

no

17. 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

18. 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

19. anonymous

Don't worry.

20. 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

21. TuringTest

thta's some crazy cartesian-polar hybrid

22. anonymous

hahaha

23. amistre64

but its good to see you thinking outside the mathical box

24. anonymous

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

25. anonymous

thank you all for the help!

26. TuringTest

welcome :)