Surface integral of vector fields (sphere)

[JulienB]

Homework Statement

Hi everybody! I'm currently training at surface integrals of vector fields, and I'd like to check if my results are correct AND if there is any shortcut possible in the method I use. I'm preparing for an exam, and I found that it takes me way too much time to solve it. I estimated that I should have around 15 mins to solve such a problem during my exam, and I'm definitely not there yet. So the problem goes:

Calculate the oriented surface integral $\int_F v \cdot dF$. The vector field is given by $v(x,y,z) = (yz,1,-xy)$, the surface $F$ is given by $F = \{ (x,y,z) \in \mathbb{R}^3 : x^2 + y^2 + z^2 = 1, y \geq 0 \}$ and the normal vector on $F$ should be oriented towards the point of origin.

Homework Equations

Surface integrals, parametrisation of surfaces, vector calculus, partal derivatives

The Attempt at a Solution

Well okay here we go:

-First I notice that the equation of the surface describes a sphere with radius $r=1$ and center $(0,0,0)$. The $y \geq 0$ indicates this is a half-sphere. So I parametrise the surface using spherical coordinates and I get:
$\phi(\varphi, \psi) = (\cos \varphi \sin \psi, \sin \varphi, \sin \psi, \cos \psi)$ with $\phi \in [0,\pi]$ (because $y \geq 0$) and $\psi \in [0,\pi]$.

-Then I take the partial derivatives with respect to $\varphi$ and $\psi$ and get:
$\partial_{\varphi} \phi (\varphi, \psi) = (- \sin \varphi \sin \psi, \cos \varphi \sin \psi, 0)$
$\partial_{\psi} \phi (\varphi, \psi) = (\cos \varphi \cos \psi, \sin \varphi \cos \psi, - \sin \psi)$

-The partial derivatives forming a tangential plane on the surface, I take the vector product of the two in order to obtain a normal vector:
$\vec{n} = \pm || \partial_{\varphi} \phi (\varphi, \psi) \times \partial_{\psi} \phi (\varphi, \psi) ||$
$= \pm (\cos \varphi \sin^2 \psi, \sin \varphi \sin^2 \psi, \cos \psi \sin \psi)$

-I could jump directly to the integral (after choosing an orientation of course), but when I see such a vector as $\vec{n}$ I prefer to normalise it otherwise I'll be trapped in the middle of many trigonometric functions when doing the integral:
$\hat{n} = \pm (\cos \varphi \sin \psi, \sin \varphi \sin \psi, \cos \psi)$

-I choose an orientation by testing the direction of say $\hat{n}_+$ for a chosen point and see if it points towards the center of the sphere or not. I chose $(\varphi, \psi) = (0,0)$ and get that for point $\phi (\varphi, \psi) = (0,0,1)$ the vector has coordinates $\hat{n}_+ = (0,0,1)$, which means it points in the opposite direction of the zero-point. Therefore, I define my normal vector as:
$\hat{n} = -(\cos \varphi \sin \psi, \sin \varphi \sin \psi, \cos \psi)$

-I can now do the integral:
$\int_F v \cdot dF = \int_S \vec{v} \cdot \hat{n} \cdot || \vec{n} ||\ d\varphi\ d\psi$
$= - \int_0^{\pi} \sin \varphi \int_0^{\pi} \sin^2 \psi\ d\psi\ d\phi$
$= - \frac{\pi}{2} \int_0^{\pi} \sin \varphi\ d\varphi$
$= -\pi$.

Is that correct? My principal doubts when doing this kind of integrals are always: did I choose the orientation right? Did I make no calculation mistake? And as I already mentioned above, I was wondering if there was a quickest way to solve those problems. This one definitely took me over half a hour, and I didn't really get stuck or anything. I often face more complicated ones and it can take over a hour then. Maybe it's just a question of practice.Thank you very much in advance for your reviews, remarks and suggestions. :)Julien.

 

[BvU]

I see you do a lot of work in 3D, but oversee that for any vector $\ \vec a = (x,y,z) \$ the inner product $\ \vec a \cdot \vec v = y$. Couldn't that shorten the time to solve ? Like, down to the time needed for multiplying the area $2\pi$ with the number 1 ?

(a bit devious, but: I'm glad our results differ, so perhaps someone will put me right ! )

[edit] the 'number' isn't 1, it's 'y', so I'm the one who is wrong...

 

[BvU]

Your notation is a bit confusing for me with $\phi$ and $\varphi$. Never mind.

From symmetry I expect the integral is zero.
Doing the work I have $| dF | = \sin\psi \;d\phi \;d\psi$ and $y = \sin\phi\cos\psi$ so a $\sin 2\psi$ appears and that gets $\int d\psi = 0$.

(a double wrong doesn't make a right, so I'll wait for confirmation).

 

[JulienB]

Hi @BvU and thanks for your answer. Yes I have seen that most people write the parameter function as $\vec{r} (u,v)$. I also like it better, but my teacher must be in love with greek letters since the notation $\phi (\varphi, \psi)$ comes from him. I'm keeping it until the exam to not get confused, but I'll drop it directly afterwards since I generally prefer to use the latin alphabet to define functions.

Mm I have trouble with your second post seeing how you arrived there. Would you mind detailing a bit? Even if it ends up being wrong, I like to see how other people think and work.

Thanks a lot! :)Julien.

 

[BvU]

second post seeing how you arrived there

So you have $$\int\limits_F \vec v\cdot\ \vec {dF}$$and $\vec v \cdot\vec x = y$.

Working in spherical coordinates is sensible, so that comes down to finding the y coordinate of $\vec dF$ in spherical coordinates.

$\vec {dF} = -\hat r \; \sin\psi \;d\psi\;d\varphi$ a vector pointing inwards along $\hat r$ with a magnitude $\sin\psi \;d\psi\;d\varphi$ (the area of a chunk $\;d\psi\;d\varphi$ of the unit sphere where r² = 1).

Its y-component is $$ -\sin\varphi\cos\psi\ \sin\psi \;d\psi\;d\varphi = - {1\over 2} \sin \varphi \sin 2\psi \;d\psi\;d\varphi$$ and the integration reads $$\int\limits_F\vec v\cdot\vec {dF} = - {1\over 2} \int\limits_{\varphi = 0}^{\pi\over 2} \int\limits_{\psi = 0}^{\pi\over 2} \sin \varphi \sin 2\psi \;d\psi\;d\varphi$$

Was that the idea ?