# Integrate this without gauss theorem

1. Mar 1, 2005

### TheDestroyer

Hi guys, I want to solve this integral without using gauss theorem to convert from double to triple integral,

my problem here is to find a general way for setting the differential surface part (ds) from the integral

if $$\vec{F}$$ was a vector field defined as:

$$\vec{F} = x^3\vec{i} + y^3\vec{j} + z^2\vec{k}$$

$$\vec{i},\vec{j},\vec{k}$$ Unit Vectors for axes $$x,y,z$$

Evaluate:
$$\oint_{s} \vec{F}\cdot\vec{ds}$$

While s is the area of the volume bounded with equation surfaces:

$$z=0, z=1-(x^2+y^2)$$

the answer will be $$\frac{5\pi}{6}$$

$$\int\int\int_{V}\vec{\nabla}\cdot\vec{F}\cdot dV$$

2. Mar 1, 2005

### dextercioby

1.Nabla has no sign on top of it.
2.Treat the flux as a genuine surface integral.Find the normal field to the surface,namely $\vec{n}$.
3.Compute the scalar product.
4.Integrate by tranforming into a double integral...

Daniel.

3. Mar 1, 2005

### Galileo

Find a parametrization of the surface.
Since the surface is given as z=f(x,y), there's an easy way to do it.
Take the projection of the surface onto the xy-plane (a circle) as the domain. Then a parametrization is (going cilindrical ofcourse):

$\vec r_r = r\hat r, \qquad \vec r_{\theta}= \theta \hat \theta, \qquad 0\leq r \leq 1, \quad \vec r_z = 1-r^2 \qquad \quad 0\leq \theta < 2\pi, \quad 0$.

Then use whatever general method you learned.

Last edited: Mar 1, 2005
4. Mar 1, 2005

### HallsofIvy

Staff Emeritus
Mildly messy but nothing very difficult.

We can write the surface as f(x,y,z)= x2+ y2+ z= 1 and so the gradient of that: grad f= 2xi+ 2yj+ k is perpendicular to the surface and we can write $d \vec s= (2xi+ 2yj+ 2zk)dA$ where dA is the differential of area in the xy-plane.
Then $$\int F \dot d \vec s= \int (2x^4 i+ 2y^4 j+ z^2 k)dA$$.

Since z= 1- (x2+ y[/sup]2[/sup]) on the surface, z2= 1- 2(x2+ y2)+ (x2+ y2)2 and the integral reduces to $$\int (3x^4+ 3y^4+ 2xy- 2x^2- 2y^2+ 1) dA[/itex]. Since the projection of the surface into the xy-plane is the unit disk, I would be inclined to convert that to polar coordinates to integrate. 5. Mar 1, 2005 ### TheDestroyer dextercioboy, it seems you didn't read what i've written, the problem is how to set the surface differential unit, after doing the dot product what shall i do with ds? and antoehr thing, Nabla has the vector on the top of it because it's a vector and has unit vectors in the definition of it, HallSofIvy, I'm sorry to tell you your solution is not correct because dA is multiplied by the unit vector of the surface gradient, not with the gradient only, right? Thank you guyz, I'll try projection on a plane, and i'll reply back, visit this post again, and thanks for the help 6. Mar 1, 2005 ### TheDestroyer Guys, I just rememebered, Projection on a plane is not a correct way, because the projection must be on XOY, XOZ, YOZ Like this (instead of multiplying with gradient unit vector): [tex]\int \vec{F}\cdot \vec{ds} = \int{F_{x}}\cdot ds_{x} + \int{F_{y}}\cdot ds_{y} + \int{F_{z}}\cdot ds_{z}$$

and this will equal by the definition of surface vector:

$$\int \vec{F}\cdot \vec{ds} = \int{F_{x}}\cdot dydz + \int{F_{y}}\cdot dxdz + \int{F_{z}}\cdot dxdy$$

and i tried using the Jacobian on those for the transformation, and that was not successful,

I'm getting the answer $$\frac{1}{3}\pi$$

I want to ask now, should I take the parameters:

$$x=r\cos\phi$$
$$dx=\cos\phi\cdot dr - r\sin\phi\cdot d\phi$$
$$y=r\sin\phi$$
$$dy=\sin\phi\cdot dr + r\cos\phi\cdot d\phi$$
$$z=1-r^2$$
$$dz=-2r\cdot dr$$

I got this by : $$df(r,\phi) = \frac{\partial f}{\partial r}dr + \frac{\partial f}{\partial \phi}d\phi$$
And multiply the differential units? i don't think this is possible !! Is it?

Please do not give any more ideas before solving this and getting the answer
$$\frac{5}{6}\pi$$

And thanks for the help

7. Mar 2, 2005

### TheDestroyer

Why No Body Is Answering? Can I Say Everybody Failed?