Why Are the Terms Squared After Substitution in Green's Theorem Integral?

[damndamnboi]

i would like to find the area bounded by the curve



(((x^2)/(a^2))+((y^2)/(b^2)))=xy/(c^2)



i used the substitution given x=(ar)cos(theta) and y=(ar)sin(theta)



i get :

(r^2cos^2(theta)+r^2sin^2(theta))^2=xy/(c^2)

thus r^4=xy/(c^2)
substituting x=(ar)cos(theta) and y=(ar)sin(theta) on the right hand side, i get
r^4=(r^2)(ab(cos<theta>)(sin<theta>)/c^2

then r^2=ab(cos<theta>)(sin<theta>)/c^2


then i used jacobian to transform dxdy to drd(theta):

i get abr(dr)(d(theta))

then i carried out the double integral
-- --
/ /
/ / abr(dr)(d(theta))
-- --

but i get 0. please advice

 

[Brad Barker]

image not working for me.

i'd recommend putting in a little bit of time to learn what you need to about latex to be able to post your problem. knowing latex is important if you intend on publishing research papers, anyway.

 

[damndamnboi]

image not working for me.

i'd recommend putting in a little bit of time to learn what you need to about latex to be able to post your problem. knowing latex is important if you intend on publishing research papers, anyway.

thx for telling me about the image not working, i have posted the question in typed form, please take a look. thx.

 

[Brad Barker]

thx for telling me about the image not working, i have posted the question in typed form, please take a look. thx.

i just skimmed your work and noticed that you had

$$dxdy = drd\theta$$.

the correct relationship is$$dxdy = rdrd\theta$$.

 

[Brad Barker]

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{xy}{c^2} \smallskip \mbox{let} x=ar\cos\theta \mbox{and} y=ar\sin\theta \smallskip (r^2\cos^2\theta+r^2\sin^2\theta)^2=\frac{xy}{c^2}$$

...wait a minute, why does your first equation not have the terms squared, but then after the substitution, they are squared again?