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

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.

• Could you show an attempt? Helping at homework is endorsed on this site, but helping homework for people who show no effort to solve their problem is not. I suggest drawing a picture – Buraian May 7 at 20:59
• I did. I had to scan my work. – SoleCore May 7 at 21:06
• You need to compute the flux of $\text{curl}\,\vec F$ across a surface, so you need to parametrize that portion of the paraboloid and use the formula to calculate a surface integral. – Ted Shifrin May 7 at 21:07
• Check your curl again... – vb628 May 7 at 21:08
• I have edited the question for searchability – Buraian May 7 at 21:21

## 3 Answers

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

• Thanks. I will try that out right now. I just learned about Stokes' Theorem and the concept isn't completely clear to me yet. – SoleCore May 7 at 21:14
• There is a book called Div , Grad curl and all that, it has one of the best explanation on this IMO @SoleCore (Also there is Khan academy but it has no problems) – Buraian May 7 at 21:15
• I also have edited my answer to put a more elaborate explanation on how to do it the right way, but more important for you is to understand the conceptual mistake you had made in the first way you solved it. – Buraian May 7 at 21:16
• Thanks a lot! I will buy that book soon too thanks! – SoleCore May 7 at 21:18
• Thanks. I feel like the book I use omits a lot of concepts I should have. I really want to fill in those gaps. If you have any other book recommendations. Please let me know! – SoleCore May 7 at 21:35

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 $$S$$.

• Yeah, I was evaluating the surface integral of curl F, but It wasn't coming out right, let me scan my homework. – SoleCore May 7 at 20:55

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

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

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

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

Convert to polar:

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

Integrating first by $$\theta.$$

$$\int_0^1 -r^3 - r^2 - 2r^2 \sqrt{1-r^2} \ dr$$

And can you take it home?