Line integral with green's theorm

In summary: i double checked my work and it looks like you may have mistyped the substitution. the substitution should have been:(r^2\cos^2\theta+r^2\sin^2\theta)^2=\frac{(x^2)/(a^2)+(y^2)/(b^2)}{c^2}
  • #1
damndamnboi
5
0
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
 
Last edited:
Physics news on Phys.org
  • #2
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.
 
  • #3
Brad Barker said:
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.
 
  • #4
damndamnboi said:
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

[tex]

dxdy = drd\theta

[/tex].

the correct relationship is[tex]

dxdy = rdrd\theta

[/tex].
 
  • #5
[tex]
\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}
[/tex]

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

Related to Line integral with green's theorm

1. What is a line integral with Green's theorem?

A line integral with Green's theorem is a mathematical tool used to calculate the area under a curve in the xy-plane. It relates the line integral of a vector field to the double integral over a region in the xy-plane.

2. What is the significance of Green's theorem in science?

Green's theorem is significant in science because it allows us to relate a line integral, which is a one-dimensional concept, to a double integral, which is a two-dimensional concept. This makes it easier to perform calculations and solve problems in various fields of science, such as physics, engineering, and mathematics.

3. How is Green's theorem applied in real-world situations?

Green's theorem can be applied in real-world situations to calculate the work done by a force, the flow of a fluid, or the amount of electrical current flowing in a circuit. It can also be used to calculate the area of irregular shapes or the mass of a thin plate.

4. What are the conditions for using Green's theorem?

The conditions for using Green's theorem are that the region in the xy-plane must be simply connected, meaning it has no holes or gaps, and the vector field must be continuous and have continuous partial derivatives in that region.

5. Can Green's theorem be used in three-dimensional spaces?

No, Green's theorem can only be used in two-dimensional spaces, specifically in the xy-plane. For three-dimensional spaces, Stokes' theorem or the Divergence theorem would need to be used instead.

Similar threads

  • Advanced Physics Homework Help
Replies
1
Views
647
Replies
6
Views
1K
  • Advanced Physics Homework Help
Replies
2
Views
875
  • Advanced Physics Homework Help
Replies
4
Views
694
  • Advanced Physics Homework Help
Replies
10
Views
2K
  • Advanced Physics Homework Help
Replies
2
Views
812
  • Advanced Physics Homework Help
Replies
0
Views
640
  • Advanced Physics Homework Help
Replies
1
Views
823
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
9
Views
2K
Back
Top