Using Stokes theorem to find the line integral over the boundary of a paraboloid in the first octant opening downward the z-axis

Question

I've been trying at this problem on my homework, but I think I am going about it the wrong way.

I tried breaking it down into the line integrals of the boundaries of the surface, but I think I might have the wrong idea about how Stokes' Theorem works.

Can someone please give me a step by step solution to this problem?

Edit

I will include my work here:

I tried several different things and none of them turned out right.

Answer 1

Your idea doesn't work because 2-d Stoke's theorem is meant for closed loops, the segments you have in each plane are NOT closed loops.

To make it work, you need to connect the segments on the y-z , x-y and z-x plane and make the whole loop and convert that line integral into a surface integral.

Here is how I'd do it, first I would find the projection of surface area of paraboloid onto the x-y plane. To do that, I set $z=0$ in that equation giving me $x^2 +y^2=1$ to see the 'domain of projection' or whatever one would call it. Now, all I have to do is plug everything into the surface integral:

$$ \int_{S} \text{curl}\vec{F} \cdot \hat{n} dS$$

I assume you know how to compute the curl (just see video on YouTube on how to do it in cartesian coordinate if you don't , it is really simple). After that I will write dS in terms of projected area:

$$ dS = \frac{dA}{\hat{n} \cdot k}$$

Since our surface is $z=1-x^2 -y^2$, I find that the unit normal is given as: $\frac{<2x,2y,1>}{\sqrt{4(x^2 +y^2)+1}}$, After this I plug everything back:

$$\int\int_{x^2 +y^2=1} \text{curl} \vec{F} \cdot \hat{n} \frac{dA}{\hat{n} \cdot \hat{k} }$$

Now this is a simple integral, I think it maybe easier to solve via the change of variables into polar coordinates

Note: We only want the integral first quadrant since the surface integral is over the first octant

Answer 2

The Stoke's Theorem says that

$$ \oint_{\partial S} \vec{F}\cdot \text{d}\vec{r} = \iint_S (\nabla \times \vec{F})\cdot \text{d}\vec{S} \text{ .} $$

If you simply evaluate the line integral by computing the integral of the pullback

$$\oint_{\partial S} \vec{F}\cdot \text{d}\vec{r} = \int_{\gamma} \vec{F}\cdot\vec{v}\: \text{d}t $$

for some curve $\vec{v} = \gamma'(t)$ then you will be ignoring the Stoke's Theorem and thus not answering the question in the correct way. You simply need to evaluate $$\iint_S (\nabla \times \vec{F})\cdot \text{d}\vec{S}$$

for your field and for the area of your closed curve $S$.

Answer 3

$\int_C F\cdot dr = \iint_S \nabla \times F\ dS$

$\nabla \times F = (-y,-z,-x)$

$dS = (-\frac {\partial z}{\partial x},-\frac {\partial z}{\partial y}, 1) = (2x, 2y,1) dx\ dy$

$\iint (-y,-z,-x)\cdot(2x, 2y, 1)\ dx\ dy\\ \iint (-2xy - 2yz - x) \ dx\ dy$

Convert to polar:

$\int_0^\frac {\pi}{2}\int_0^1 (-2r^2\cos\theta\sin\theta - 2r\sin\theta(1-r^2) - r\cos\theta) (r \ dr\ d\theta)$

Integrating first by $\theta.$

$\int_0^1 -r^3 - r^2 - 2r^2(1-r^2) \ dr$

And can you take it home?