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.

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

1 Answer 1

  • 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

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


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

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 $C(0)=(-2,-1)$.

  • $\begingroup$ 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. $\endgroup$
    – Artemisia
    Nov 20, 2013 at 10:13
  • 1
    $\begingroup$ I edit my answer, just a sec. $\endgroup$
    – Avitus
    Nov 20, 2013 at 10:23
  • $\begingroup$ Thanks a ton! :) This is much easier than involving the parameterization $\endgroup$
    – Artemisia
    Nov 20, 2013 at 10:31
  • 1
    $\begingroup$ 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! $\endgroup$
    – Avitus
    Nov 20, 2013 at 11:10
  • 1
    $\begingroup$ you are right: I shortened notation too much in the end. I modified it: thank you again! $\endgroup$
    – Avitus
    Nov 21, 2013 at 15:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .