How can I evaluate this line integral directly?

Question

Let $P = (x-1)y^{2}$ and $Q = (y+1)x^{2}$ How can I evaluate $\oint_C Pdx + Qdy$ without simplifying using Green's theorem.

Counterclockwise around the triangle defined by the points: $(0,0) (0,1) (\frac{1}{2}, 0)$

How could I reduce the integral to either in only dx or dy, and is that even necessary?

Thanks very much in advance

Answer 1

If $C = C_1+C_2 + \cdots + C_N$ then we have;

$$\oint_C f = \sum_{j=1}^N \oint_{C_j} f$$

Now just remember to parametrize a line segment use $p(1-t)+tq$ where $t \in [0,1]$.

Answer 2

Using Green's Theorem you can say $$ \oint_C P dx + Q dy = \iint_D \left( \frac{dQ}{dx} - \frac{dP}{dy}\right) \ dxdy = \iint_D (2yx +2x -2xy +2y) \ dx dy $$ Where $C = \partial D$. You can therefore compute : $$ 2\cdot \int_0^{1/2} dx \int_0^{1-2x} dy\ (x+y) $$

Answer 3

Line from $\;(0,0)\to (0,1)\;$ :

$$\ell_1:\;t(0,1)+(1-t)(0,0)=(0,t)\;,\;\;t\in [0,1]\implies$$

$$\int_{\ell_1}Pdx+Qdy=\int_0^1\left((-1)t^2(0\cdot )+(t+1)0^2\right)dt=0$$$${}$$

Line from $\;(0,1)\to \left(\frac12,0\right)\;$ :

$$\ell_2:\;t\left(\frac12,0\right)+(1-t)(0,1)=\left(\frac t2,1-t\right)\;,\;\;t\in [0,1]\implies$$

$$\int_{\ell_2}Pdx+Qdy=\int_0^1\left(\left(\frac t2-1\right)(1-t)^2\cdot\frac12dt+(2-t)\frac{t^2}4(-dt)\right)=$$

$$=\frac14\int_0^1(t-2)(t-1)^2dt+\frac14\int_0^1(t^3-2t^2)dt=\frac14\int_0^1\left(2t^3-6t^2+5t-2\right)dt=$$$${}$$

$$=\frac14\left(\frac12-2+\frac52-2\right)=-\frac14$$

Line from $\;\left(\frac12,0\right)\to (0,0)\;$ :

$$\ell_3:\;t(0,0)+(1-t)\left(\frac12,0\right)=\left(\frac12(1-t),0\right)\;,\;\;t\in [0,1]\implies$$

$$\int_{\ell_3}Pdx+Qdy=\int_0^1\left(-\frac12-t\right)\left(-\frac12\right)+\left(1)0^2\right)dt=\int_0^1\left(\frac14+\frac t2\right)dt=$$

$$=\frac14+\frac14=\frac12$$

so the line integral equals $\;-\cfrac14+\cfrac12=\cfrac14\;$

Using Green's theorem:

$$\frac{\partial Q}{\partial x}=2x(y+1)\;,\;\;\frac{\partial P}{\partial y}=2(x-1)y\implies\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=2(x+y)$$

and the integral becomes on the given region:

$$\int_0^{1/2}\int_0^{-2x+1}2(x+y)dydx=\int_0^{1/2}\left(2x(-2x+1)+(2x-1)^2\right)dx=$$

$$=\int_0^{1/2}(-2x+1)dx=-\frac14+\frac12=\frac14$$

Answer 4

Hint:

You can express $P$ as a function of $x$ alone and $Q$ as a function of $y$ alone, by means of the equations of the line segments.

For instance, $(0,0)\to(0,1)$ is $x=0$ for $y\in[0,1]$, and

$$\oint_{(0,0)\to(0,1)} Pdx + Qdy=\oint_{(0,0)\to(0,1)}0+ (y+1)0^2dy=0.$$