# Solve the line integral without Green's formula

I'm starting on double and line integrals, and I'm stuck at this question. It asks of me to calculate the following integral without using Green's theorem. Usually with Green's theorem I use polar coordinates easily, but I can't seem to make the same substitution here.

$$\int_{L}{(2x-y)dx+(x-y)dy}$$

$$L = \{(x,y): x^{2}+y^{2}=2y, x\geq0\}\cup\{(x,y):x^{2}+y^{2}=4, x\leq0,y\geq0\}$$

The curve $$L$$ is oriented from the point $$(0,0)$$, that is counter-clockwise.

I am not sure which subsitution to use. I always used Greens theorem so far. How would I go about solving this without it? I thought of using some parameter $$t$$ but couldn't quite specify its range. Thank you.

• Tell your teacher that this is a sick way to define a $1$-chain. – Christian Blatter Nov 20 '18 at 19:08

## 1 Answer

You have two curves.

first part.

$$x = 2\sin t\cos t = \sin 2t\\ y = 2\sin^2 t = 1-\cos 2t$$

$$\int_\limits{0}^{\frac {\pi}{2}} (2\sin 2t + 1 - \cos 2t )(2\cos 2t)+(\sin 2t +1-\cos 2t)(2\sin 2t) \ dt$$

and the second part

$$x = 2\cos t\\ y = 2\sin t$$

$$\int_\limits{\frac {\pi}{2}}^{\pi} (4\cos t - 2\sin t)(-2\sin t)+(2\cos t - \sin t)(2\cos t) \ dt$$

If you wanted to use greens.

Then the you would add the straight line from $$(-2,0)$$ to $$(0,0)$$ to close the contour.

Then

$$\int_{C_1} P(x,y)\ dx + Q(x,y)\ dy + \int_{C_2} P(x,y)\ dx + Q(x,y)\ dy = \iint \frac {\partial P}{\partial y} - \frac{\partial Q}{\partial x}\ dA\\ \int_{C_1} P(x,y)\ dx + Q(x,y)\ dy = \iint \frac {\partial P}{\partial y} - \frac{\partial Q}{\partial x}\ dA- \int_{C_2} P(x,y)\ dx + Q(x,y)\ dy$$

• Thanks, I understand everything except how you got the first parts parametric equations. Since the first part of the curve is the right side of a circle of radius 1 and y offset of 1, shouldn't y be $y=1+sin(t)$. And $x=cos(t)$. – math101 Nov 20 '18 at 22:32
• You could use that parameterization, it would be for a different range of t. That is fine. I said, $x = r\cos t, y = r\sin t$ plugging into $x^2 + y^2 = 2y$ gives $r = 2\sin t$ And put that back in for $r$ in the equations of $x,y$ – Doug M Nov 20 '18 at 22:42
• So that's it! I see now , thanks! – math101 Nov 20 '18 at 23:13