# Thread: Stokes formula and the curvilinear integral of the second kind

1. ## Stokes formula and the curvilinear integral of the second kind

Hello all!
I can not solve a difficult task from the vector analysis:
Using Stokes formula, calculate the curvilinear integral of the second kind: $\displaystyle I = \int\limits_{OA} {yzdx + 3xzdy + 2xydz,}$
where $\displaystyle OA = OB \cup BC \cup CA$, curve $\displaystyle x = t\cos t,{\text{ }}y = t\sin t,{\text{ }}z = {t^2},{\text{ }}0 \leqslant t \leqslant 2\pi ,{\text{ }}O = \left( {0,0,0} \right),{\text{ }}A = \left( {2\pi ,0,4{\pi ^2}} \right).$

I have two carefully (I thought) addressed this problem , but my answer is clearly not the same as the right one.

2. Originally Posted by DeMath
Hello all!
I can not solve a difficult task from the vector analysis:
Using Stokes formula, calculate the curvilinear integral of the second kind: $\displaystyle I = \int\limits_{OA} {yzdx + 3xzdy + 2xydz,}$
where $\displaystyle OA = OB \cup BC \cup CA$, curve $\displaystyle x = t\cos t,{\text{ }}y = t\sin t,{\text{ }}z = {t^2},{\text{ }}0 \leqslant t \leqslant 2\pi ,{\text{ }}O = \left( {0,0,0} \right),{\text{ }}A = \left( {2\pi ,0,4{\pi ^2}} \right).$

I have two carefully (I thought) addressed this problem , but my answer is clearly not the same as the right one.
I'm not sure what your B and C is but I'm going to guess.

The surface whose boundary is $\displaystyle x = t \cos t,\; y = t \sin t,\; z = t^2$ is given parametrically as

$\displaystyle \vec{r}(s,t) = <s \,t \cos t,\; s \, t \sin t,\; t^2>$

where $\displaystyle 0 \le s \le 1$ (see attached graph). The surface is blue, your boundary is in black and I think the B and C you mention are the red lines in the graph one horizontal and one vertical. Now the line integral is easy. Introducing your parametric equations directly into the line integral gives

$\displaystyle I = \frac{32 \pi^5}{5}$

Now for the Stoke's part. If $\displaystyle {\bf F} = <yz, 3xz, 2xy>$ so $\displaystyle \nabla \times {\bf F} = <-x, -y, 2z> = < - s \,t \cos t, - s \, t \sin t, 2t^2>$

Further $\displaystyle \vec{n} = \vec{r}_s \times \vec{r}_t = <-2 t^2 \sin t, 2 t^2 \cos t, s t^2>$

so that $\displaystyle \iint_s \nabla \times {\bf F} \cdot \vec{n} dS = \int_0^{2\pi} \int_0^1 2s t^4 ds dt = \frac{32 \pi^5}{5}$

3. If I understand correctly, the non-closed curve $\displaystyle OA = OB \cup BC \cup CA$ lies on the surface of paraboloid $\displaystyle z = {x^2}+{y^2}$.
I started to solve this problem by adding a curve of the integration to a closed contour $\displaystyle G$ of the curve $\displaystyle z = {x^2}$, lying in the plane $\displaystyle XZ$. But I was mistaken in the computation, so my answer is very different from the correct answer $\displaystyle \frac{{32}}{5}{\pi ^5}$.

Many thanks, Prof. Danny!

4. If I understood correctly (?), then this task can be interpreted graphically as follows.

5. Originally Posted by DeMath
If I understood correctly (?), then this task can be interpreted graphically as follows.

Then all we need is a different parametization of the surface. In this case

$\displaystyle \vec{r} = < s \cos t, s \sin t, s^2>$
so

$\displaystyle \vec{r}_u \times \vec{r}_v = <-2s^2 \cos t, - 2 s^2 \sin t, s>$

so $\displaystyle \nabla \times {\bf F} \cdot \vec{n} = 4s^3$

and $\displaystyle \iint_S \nabla \times {\bf F} \cdot \vec{n} dS = \int_0^{2 \pi} \int _0^t 4s^3 ds\,dt$ which gives the same answer.