In the problem below, parametrize the plane curves below in such a way that it traversed only once , its unit normal vector towards the interior of the bounded region it enclose.
Then use Green theorem to evaluate line integral:
$$ \int_C y^2dx+xydy $$
where $ \ C \ $ is the boundary curve of the region lying between $ \ y=0 , \ y=\sqrt x \ $ and $ \ x=9 \ $
Answer:
Let $ \ y=t, \ x=t^2 \ , \ 0 \leq t \leq 3 \ $ be the parametrization.
Then,
$ F(x,y)=y^2 \hat i+xy \hat j, \\ r(t)=x \hat i+y \hat j=t \hat i+t^2 \hat j \\ \Rightarrow dr=(\hat i+2t \hat j)dt \ \ and \ \ F(r(t))=t^2 \hat i+t^3 \hat j $
Thus,
$ \int_C y^2 dx+xy dy =\int_{0}^3 \ F(r) \cdot dr = \int_0^3 (t^2+2t^4) dt = 9+2 \cdot \frac{3^5}{5}$
Am I right ?
Help me
Greens Theorem states that $$\int_C y^2 dx + xy dy = \iint_D \frac d {dx} xy - \frac d {dy} y^2$$ where $D$ is the region enclosed by $C$. This can be simplified to $$= \iint_D y - 2y = \iint_D -y$$ It seems to me the problem wishes you to use Greens Theorem by evaluating this much simpler integral to get the integral over $C$
You seemed to try to evaluate the integral without using Greens Theorem which will get you the right answer too, however there are a couple errors:
You swapped the variables in $r(t)$. It should be that $r(t) = x \hat i + y \hat j = t^2 \hat i + t \hat j$. Thus $dr = 2t \hat i + \hat j$ and what you end up integrating is $\int_0^3 2t^3 + t^3\ dt$
You forgot to integrate over the $x=9$ and $y = 0$ portions of $C$.