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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.

Artemisia
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1 Answers1

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  • 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 $C(0)=(-2,-1)$.

Avitus
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  • 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. – Artemisia Nov 20 '13 at 10:13
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    I edit my answer, just a sec. – Avitus Nov 20 '13 at 10:23
  • Thanks a ton! :) This is much easier than involving the parameterization – Artemisia Nov 20 '13 at 10:31
  • Apparently this answer is wrong. I don't see the mistake. – Artemisia Nov 20 '13 at 10:37
  • Me too: do you see any typo? Which is the answer in your textbook? – Avitus Nov 20 '13 at 11:03
  • It is an online question bank. No answer provided until we enter the right one. – Artemisia Nov 20 '13 at 11:03
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    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! – Avitus Nov 20 '13 at 11:10
  • I tried to solve this sum using the method you explained (thank you so much for it :) ) and I got the same answer as you did. The numerical value is $1.20$ but the system says it's incorrect. – Artemisia Nov 20 '13 at 11:14
  • I have a question in the final step. You have substituted for $\varphi (1)$ and $\varphi (0)$. However, what are the values of $x$ and $y$, for $\varphi (x,y)$ is a function of both $x$ and $y$? Could that be the error? – Artemisia Nov 21 '13 at 14:34
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    you are right: I shortened notation too much in the end. I modified it: thank you again! – Avitus Nov 21 '13 at 15:44
  • Haha thank you! This question was such a pain! :D – Artemisia Nov 21 '13 at 15:47