Line integrals of given curves

Question

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.

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

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

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