First of all, sorry for the sketch.. I would be glad if you show me another ways to improve this drawn.
I'm studying Line integrals, and this question is really boring me. Please, can someone put the steps of the resolution WITHOUT the Green's theorem, and parameterizing in $t$ variable?
The question:
Let $C$ be the oriented curve (see the image). Evaluate $$\int_C (x^3 y^2 + y)dx +xdy$$
The answer must be $-1/8$
Image:
Your integral is sum $$ I = I_1 + I_2 + I_3 = \int\limits_{y=x^2} + \int\limits_{y=1} + \int\limits_{x=0} $$ Ok, let's evaluate. $$ I_1 = \int\limits_{y=x^2} (x^3y^2 + y)\,dx + x\,dy = \int\limits_{x=0}^{x=1} (x^3(x^2)^2 + x^2)\,dx + x\cdot2x\,dx =\\= \int\limits_{x=0}^{x=1} (x^7 + 3x^2)\,dx = \frac98 $$ $$ I_2 = \int\limits_{y=1} (x^3y^2 + y)\,dx + x\,dy = \int\limits_{x=1}^{x=0} (x^3 + 1)\,dx = -\frac54 $$ $$ I_3 = \int\limits_{x=0} (x^3y^2 + y)\,dx + x\,dy = 0 $$ Now $$ I = I_1 + I_2 + I_3 = \frac98 - \frac54 = -\frac18. $$
We could simplify this integral a little: $$ \oint (x^3y^2 + y)\,dx + x\,dy = \oint x^3y^2\,dx + \oint y\,dx + x\,dy = \oint x^3y^2\,dx + \oint d(xy) = \oint x^3y^2\,dx $$
Or to use Green: $$ I = \iint \left(\frac{\partial(x^3y^2 + y)}{\partial y} - \frac{\partial(x)}{\partial x}\right)dx\,dy = \iint 2x^3y\,dx\,dy = \int_0^1 x^3\,dx \int\limits_{y=x^2}^{y=1} 2y\,dy =\\= \int_0^1 x^3(x^4 - 1)\,dx =\frac18-\frac14 = -\frac18 $$