# Line integrals of given curves

This question has an integral $$\int(x^4+4xy^3)dx+(6x^2y^2-5y^4)dy$$to be evaluated on the parametric curve $$C:(-(t+2)\cos(\pi t^2), t-1)$$I took the partial derivatives of the terms in the bracket and subtracted them to get $0$. However, this is not the right answer. I don't know any other method to solve such integrals.

• Note $dx=\frac{dx}{dt}dt$ Nov 20, 2013 at 9:04
• What is the initial point and terminal point of the curve?
– Paul
Nov 20, 2013 at 9:05
• It isn't given. Only, $t$ is from the interval $[0,1]$. Nov 20, 2013 at 9:06

• Direct approach

Let us write the integral as

$$\int_C F\cdot dC:=\int_0^1 F(C(t))\cdot\frac{dC}{dt}dt,$$

with $F(x,y):=(F_1(x,y),F_2(x,y))=(x^4+4xy^3,6x^2y^2-5y^4)$ and $C:[0,1]\rightarrow R^2$, with $C(t):=(-(t+2)\cos(\pi t^2), t-1)$. The integral is really complicated and we do not want to perform all computations.

• Searching for a potential $\varphi(x,y)$

Let us try another way, i.e. let us have a deeper look at the original formulation of our integral:

$$\int_C F\cdot dC:=\int F_1dx+F_2dy$$

If we could find a $C^1$ function $\varphi(x,y)$ s.t.

$$F_1:=\frac{\partial \varphi}{\partial x},$$ $$F_2:=\frac{\partial \varphi}{\partial y},$$

then our integral would be equal to

$$\int_C F\cdot dC:=\int_C \frac{\partial \varphi}{\partial x}dx+\frac{\partial \varphi}{\partial y}dy=(\text{using the definition of the integral along a curve})= \int_0^1\frac{d\varphi(C_1(t),C_2(t))}{dt}dt=\varphi(C_1(1),C_2(1))-\varphi(C_1(0),C_2(0)).$$

This proof, if it not clear, can be found on all textbooks on Analysis.

To find such $\varphi$, if it exists, we must solve the equations

$$x^4+4xy^3=\frac{\partial \varphi}{\partial x},$$ $$6x^2y^2-5y^4=\frac{\partial \varphi}{\partial y}.$$

Let us solve the first equation; we arrive at

$$x^4+4xy^3=\frac{\partial \varphi}{\partial x}\Rightarrow \varphi(x,y)=\frac{x^5}{5}+2x^2y^3+\rho(y),$$

for some function $\rho=\rho(y)$. Plugging the above $\varphi(x,y)$ in the second equation we arrive at

$$\frac{\partial }{\partial y}\left(\frac{x^5}{5}+2x^2y^3+\rho(y)\right)\stackrel{!}{=} 6x^2y^2-5y^4,$$

i.e.

$$6x^2y^2+\frac{d\rho }{d y}\stackrel{!}{=} 6x^2y^2-5y^4,$$

which implies $\frac{d\rho }{d y}=-5y^4$, or $\rho(y)=-y^5$.

In summary, the potential function is given by

$$\varphi(x,y)=\frac{x^5}{5}+2x^2y^3-y^5,$$

and the original integral is

$$\int_C F\cdot dC=\varphi(3,0)-\varphi(-2,-1).$$

as $C(1)=(3,0)$ and $C(0)=(-2,-1)$.

• I found $\frac {\partial\varphi}{\partial y}$ to be $2x^2y^3-y^5+\rho(x)=\varphi(x,y)$. I am not sure of how to proceed... because I am confused between the limits and the terms to be integrated. Nov 20, 2013 at 10:13
• I edit my answer, just a sec. Nov 20, 2013 at 10:23
• Thanks a ton! :) This is much easier than involving the parameterization Nov 20, 2013 at 10:31
• Then we cannot infer anything about the answer :-) Apart from the numerical value (I will check it again), please be sure of the "method". It uses the theory of conservative vector fields and potential functions. Regards! Nov 20, 2013 at 11:10
• you are right: I shortened notation too much in the end. I modified it: thank you again! Nov 21, 2013 at 15:44