$$\underset{S}{\iint} (yz\text{dx}\text{dy}+xy\text{dxdz}+xz\text{dydz})$$

Where S is the outer side of a part of a cylinder: $ x^2+y^2 = r^2 ; x\le0,\ y\ge0, \ 0\le z\le H$

The problem about that is - I have tried parsing this in few different ways through Wolfram Mathematica 10, like converting [x,y,z] to cylindrical and pasting it instead of (x , z) , but then ouput was just a volume of cylinder formula followed with the expression above in altered form. I am kinda new to the app, to be honest.

CoordinateTransform["Cylindrical" -> "Cartesian" , {x, y, z}]
{x Cos[y], x Sin[y], z}
f[x_ y_ z_] := (y*z/dz) + (x*y/dy) + (z*x/dx)
integrate[f[x Cos[y], x Sin[y], z] , {y, 0, 2*pi} , {z, 0, H}]

Output: integrate[f[x Cos[y], x Sin[y], z], {y, 0, 2 pi}, {z, 0, H}]

Weird? Well, at least for me, I might be bad at Mathematica language. Then, I tried to rather not struggle around making a definition of f(x_ y_ z_) and be more straightforward. (Pasting a screenshot due to troubles copying that)enter image description here Next I have found out Integrate is written with capital and tried first way again.

f[x_ y_ z_] := (y*z/dz) + (x*y/dy) +(z*x/dx)
Integrate[f[x Cos[y],x Sin[y],z] , {y,0,2*pi} , {z,0,H}]

Out: $$\int _0^{2 \text{pi}}\int _0^Hf(x \cos (y),x \sin (y),z)dzdy$$ Then:

Integrate[f[x Cos[y], x Sin[y], z] , {x, 0, r}, {y, 0, 2*pi} , {z, 0, H}]

Out: $$\int _0^r\int _0^{2 \text{pi}}\int _0^Hf(x \cos (y),x \sin (y),z)dzdydx$$

BUT... As you see, no result at all... I have been thinking about integrating over Boole (but deleted that .nb) , integrating the initial expression, putting limits like

Integrate[expr , {x,y,z} \[Element] Cylinder[{{0,0,0},{0,0,h}},r] ]

And I have no out due to deleted .nb so simply put - I got the same expression, all put up as fraction of their sum by ddx ddy ddz . BUT multiplied by volume of Cylinder (very smart) As suggested, fixed definition of f:

f[x_, y_, z_] := (y*z/dz) + (x*y/dy) + (z*x/dx)
integrate[f[x Cos[y], x Sin[y], z] ,{x,0,r}, {y, 0, 2*pi} , {z, 0, H}]

Out: $$\frac{H r^2 \sin (\text{pi}) \left(3 \text{dx} \text{dy} H \sin (\text{pi})+4 \text{dx} \text{dz} r \sin (\text{pi}) \cos ^2(\text{pi})+3 \text{dy} \text{dz} H \cos (\text{pi})\right)}{6 \text{dx} \text{dy} \text{dz}}$$

So basically, I can't really find what I want with the expressions above

//I am kind of sorry for being bad at mathematica language :(

  • $\begingroup$ Show us what you have tried so far, and we'll go from there. $\endgroup$
    – MarcoB
    Dec 8 '15 at 14:13
  • $\begingroup$ Edited OP. Check this out! $\endgroup$ Dec 8 '15 at 14:44
  • $\begingroup$ The correct definition of function is f[x_, y_, z_] := ... $\endgroup$
    – ybeltukov
    Dec 8 '15 at 15:14
  • 1
    $\begingroup$ For starters, the function integrate is case sensitive, so you need it to be Integrate[f[...], ...] instead of integrate[...]. Second, you use dx, dy, and dz as variables in your function f[...], but dx, dy, and dz are variables with those names and not the same as partial-derivative notation. If you want to enter in the partial derivatives, type esc-d-d-esc in the Mathematica front-end. $\endgroup$
    – nben
    Dec 8 '15 at 20:40
  • 1
    $\begingroup$ ... you also don't define x anywhere, so the x you pass to f[...] in the integral is just a symbol, and not any particular value. Also, Pi is case sensitive in Mathematica --- 'pi' is just a symbol, Pi is the actual value. It seems that what you want to do is to use a parametric integral, Integrate[f[theta, z], {theta, 0, 2*Pi}, {z, 0, H}] where f[theta, z] needs to be defined for theta and z (rather than x, y, z). $\endgroup$
    – nben
    Dec 8 '15 at 20:44

A surface integral is often written in this form:

enter image description here

where dx[i] ^ dx[j] = - dx[j] ^ dx[i] (antikommutative)

We have to translate your integral:

v1 = x z;
v2 = -x y;
v3 = y z;
v = {v1, v2, v3};

With GAUSS we can do

div = Div[v, {x, y, z}]
-x + y + z

and with the region we obtain

reg = ImplicitRegion[
   x^2 + y^2 <= r && 0 <= z <= H && x <= 0 && y >= 0, {x, y, z}];
Integrate[ div, {x, y, z} \[Element] reg, Assumptions -> r > 0 && H > 0]
1/24 H (3 H \[Pi] + 16 Sqrt[r]) r
  • $\begingroup$ Thanks! It took me a while to figure out how to interpret it in mathematica... I'll try this later and give feedback. $\endgroup$ Dec 11 '15 at 7:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.