How to choose the right bounds for this Stokes' theorem problem

Question

There is a part of a cylinder $y^2 + z^2 = 1$ between $x =0$ and $x = 3$ and above the xy-plane. The boundary of this surface is a curve C which is oriented counter-clockwise when viewed from above. Using Stokes' theorem, compute $$ \oint_C \vec{F} \cdot d \vec{r} $$ where $\vec{F}(x,y,z)= \langle y^3, 4xy^2, yz \rangle$

Stokes' theorem: $$ \oint_C \vec{F} \cdot d \vec{r} = \int\int_S Curl \vec{F} \cdot d\vec{s} $$

I have already calculated Curl $\vec{F}$ to be $\langle z,0,y^2 \rangle$. I just do not know what bounds/how to calculate those bounds to integrate by and what $d\vec{s}$ is.

Answer 1

You have $F(x,y,z) = (y^3,4xy^2,yz)$. If we write $F=(P,Q,R)$, then Stokes' theorem says that $$\int_C F\cdot dr = \int_C P \ dx + Q \ dy + R \ dz = \int_S (\partial_y R - \partial_z Q) \ dydz + (\partial_z P - \partial_x R) \ dzdx + (\partial_x Q - \partial_y P) \ dxdy.$$ This simplifies to $$\int_S z \ dydz + y^2 \ dxdy. $$ From here you just need to parameterize and compute the above integral. Does that make sense? I suspect that approaching the problem using cylindrical coordinates oriented appropriately will help.

Edit: For the limits of integration, the first equation tells you that you want $0\leq r\leq 1$. For $\theta$, the key info is that we want the region to lie above the $xy$-plane. You should be able to take $0\leq\theta\leq\pi$. I'm on my phone, so this is all a bit off the cuff.

Answer 2

I will first show a simpler approach to calculate this, though the answer may be long due to explanations.

$\vec F = (y^3,4xy^2,yz)$
$\nabla \times \vec F = (z, 0, y^2)$

You need to find line integral of closed curve $C$ which is oriented anti-clockwise and is defined by $4$ pieces

i) curve $y^2+z^2 = 1, x = 3$ from $y = -1$ to $y = 1$
ii) line $y = 1, z= 0$ from $x = 3$ to $x = 0$
iii) curve $y^2+z^2 = 1, x = 0$ from $y = 1$ to $y = -1$
iv) line $y = -1, z= 0$ from $x = 0$ to $x = 3$

The line integral of $\vec F$ over curve $C$ is same as surface integral of curl of $\vec F$ over cylindrical surface $S: y^2 + z^2 = 1, 0 \leq x \leq 3$.

But surface integral of $\nabla \times \vec F$ over cylindrical surface can also be evaluated with help of divergence theorem. To apply divergence theorem, we close the surface and our closed surface has following surfaces.

i) Cylindrical surface $S$ defined above
ii) A rectangular disk $S1: 0 \leq x \leq 3, -1 \leq y \leq 1, z = 0$
iii) Half circular disks $S2: y^2+z^2 \leq 1$ at $x = 0$
iv) Half circular disks $S3: y^2+z^2 \leq 1$ at $x = 3$

Now we know that $\nabla \cdot (\nabla \times \vec F) = 0$. In other words, the divergence of a vector field which is curl of another vector field is zero. So flux through $4$ surfaces must cancel each another as the net flux through closed surface is zero.

For surfaces iii) and iv), please note that outward normal vectors are $(-1, 0, 0)$ and $(1, 0, 0)$. As our vector field (curl of $\vec F$) is not a function of $x$ and both disks are $y^2+z^2 \leq 1$, both these surface integrals will cancel each another. So it is clear that,

$\displaystyle \iint_{S} (\nabla \times \vec F) \cdot \hat n \ dS = - \iint_{S1} (\nabla \times \vec F) \cdot \hat n \ dS$

For rectangular disk $S1$ in $XY$ plane, outward normal vector is $(0, 0, -1)$. So,

$\displaystyle \iint_{S1} (\nabla \times \vec F) \cdot \hat n \ dS = \int_0^3 \int_{-1}^1 - y^2 \ dy \ dx = -2$

So the surface integral of $\nabla \times \vec F$ over $S$ or the line integral of $\vec F$ over curve $C$ is $\fbox{2}$

Coming back to finding surface integral over $S$ directly, cylindrical surface can be parametrized as $r(x, \theta) = (x, \cos\theta, \sin\theta), 0 \leq x \leq 3, 0 \leq \theta \leq \pi$.

$\nabla \times \vec F = (\sin\theta, 0, \cos^2\theta)$
$r_\theta \times r_x = (0, \cos\theta, \sin \theta)$

$\displaystyle \iint_{S} (\nabla \times \vec F) \cdot \hat n \ dS = \int_0^3 \int_0^{\pi} \cos^2\theta \sin\theta \ d\theta \ dx = 2$.