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Thank you very much, this gives me the correct answere :D


can i perform this check without units? or do i have to type everything without units then?



Thank you very much, maple accept both, but somehow mapel returns false for all arguments, even when im turning the inequality sign


as far as i can see they should all return true

8.982892831*Unit('mm') < 45.50833333*Unit('mm'), 4.189790007*Unit('mm') < 36.51333333*Unit('mm'), 3.913902969*Unit('mm') < 28.04666667*Unit('mm'), 3.620745836*Unit('mm') < 19.05000000*Unit('mm'), 4.482892831*Unit('mm') < 45.50833333*Unit('mm'), 4.189790007*Unit('mm') < 36.51333333*Unit('mm'), 8.793627806*Unit('mm') < 177.8000000*Unit('mm'), 3.327643012*Unit('mm') < 10.05500000*Unit('mm'), 4.189790007*Unit('mm') < 36.51333333*Unit('mm'), 3.913902969*Unit('mm') < 28.04666667*Unit('mm'), 3.913902969*Unit('mm') < 28.04666667*Unit('mm'), 3.767378711*Unit('mm') < 23.55000000*Unit('mm'), 3.913902969*Unit('mm') < 28.04666667*Unit('mm'), 3.620745836*Unit('mm') < 19.05000000*Unit('mm'), 3.620745836*Unit('mm') < 19.05000000*Unit('mm'), 3.913902969*Unit('mm') < 28.04666667*Unit('mm'), 3.913902969*Unit('mm') < 28.04666667*Unit('mm')
seq(is(tm[i] < (1/6)*Diameter[rør][i]), i = 1 .. 17);
false, false, false, false, false, false, false, false, false, 

  false, false, false, false, false, false, false, false



Thanks :) 

Your assumptions are absolutely right =) 


Hi, im sorry for not being clear, If I knew what I was doing I would probably not asked this question.

After exploring the help center i came out with this:


p := plot3d([1/(r^2+1), (1/2)*r^2], r = 0 .. 1, y = 0 .. 2*Pi)


changecoords(p, cylindrical)





Download coords.mw

Even if id like to plot straight in cylindrical this would work for now :)


At least thanks for taking time to read and respond to my thread


I wanted to share with you that this option for the plots made both spcacurve and the surface plot appeare perfect again.

I read about the issue at maplesoft support site after you led me towards the answer, I tried what was suggested but seems like my driver cannot be updated for some really strange cause. Apperantly there is more people struggling with the driver update for gpu on yoga2.

the GPU is Intelr(R) hd graphics family.


Thank you for your interest in helping me. 

I will check that as my first priority when I arrive home but wanted to check if you could address the poor plots of the surfaces as well. 

The grids are missing and the intersection does not look smooth at all. 


Im not 100 sure of my gpu but it's some sort of integrated gpu, the machine is Lenovo yoga pro 2 with 3200 x 1880 resolution scaled down to HD resolution. 


Will aftercome with results and exact specifications ASAP. 



@tomleslie A great approach to vizualise the problem!




P1 := plot3d([rhs(S1), rhs(S2)], x = -2 .. 0, y = sqrt(-x^2-2*x) .. -sqrt(-x^2-2*x), color = [red, green])


Finner parametrisering for kurven: flytter begge surfaces til LHS og setter dem lik 0 Av å til må man complete square

x^2+y^2=r^2 Parameter for x,y=r*cos(t),r*sin(t)

Bruk trig identities om du sliter. Simplify'trig'

rhs(S1)-rhs(S2); CompleteSquare(rhs(S1)-rhs(S2))





Ser at parametriseringen kan settes slik x=cos(t)-1    y=sin(t) må finne parameter for z

simplify(subs(x = cos(t)-1, y = sin(t), [S1, S2]))

[z = -2*cos(t)+2, z = -2*cos(t)+2]


R := [cos(t)-1, sin(t), -2*cos(t)+2]

[cos(t)-1, sin(t), -2*cos(t)+2]


P2 := spacecurve(R, t = 0 .. 2*Pi, thickness = 5)





Download file.mw

@Markiyan Hirnyk 

Thank you


Thanks for still having interest in my topic. Yes I know I can, but i dont know how to express the integral bounds r if the circle is arbutary placed. In this case the bounds is 0..-2cos(t) but what if the circle was centred at (-2,-1) how would i get the bound for r? 




Thank you for your detailed solution, the integrals deffinatly gives the right area. However the applications I will need this for would be in integrals where the intergrand needs to be of certain types, i.e stokes theoreme where the integral must be a double integral with region D=dA and the intergrand needs to be curl of F crossed with normal vector of the surface.

I dont disagree that this approach will be harder then other approaches, but in case my finals ask me to do it this certain way i want to be prepared and know how I can utilize maple to my advantage.

If nothing is specified and im free to choose method the VectorCalculus package with LineInt and path will my first choice for simplicity.

The problem also becomes easy when using cartesian coordinates, but in worst case scenario they will ask me to do it in polar.


This is work shown for this special case, but I don´t know how i can make it general in maple if the domain has different radius or its centered at different coordinates.


Below I will provoide the work I have done to solve a line integral using stokes theorem for a defined vectorfield.

Hope this makes what I need a little bit clearer, I know my explanation can be a bit bad since my mother tounge is not english.

Your help has been greatly apperciated so far :)


restart; with(plots); with(LinearAlgebra); with



S1 := z = -2*x

z = -2*x


S2 := z = x^2+y^2

z = x^2+y^2


plot3d([rhs(S1), rhs(S2)], y = -1 .. 1, x = -2 .. 0, color = [green, red], scaling = constrained, orientation = [55, 70, 0])


Y := solve(rhs(S1) = rhs(S2));

{x = x, y = (-x^2-2*x)^(1/2)}, {x = x, y = -(-x^2-2*x)^(1/2)}


-2, 0


plot3d([rhs(S1), rhs(S2)], y = -sqrt(-x^2-2*x) .. sqrt(-x^2-2*x), x = -2 .. 0, color = [green, red], orientation = [0, 0, 0])




Stokes surface integral


Normalvektor kalles N, vær obs på retningen.

N := [-(diff(rhs(S1), x)), -(diff(rhs(S1), y)), 1];

[2, 0, 1]


Nå blir integralet ett areal integral projisert på x-y planet NdS=[-Zx,-Zy,1]dA

F := [cos(z), x^2, 2*y];

[cos(z), x^2, 2*y]


CurlF := [diff(F[3], y)-(diff(F[2], z)), diff(F[3], x)-(diff(F[1], z)), diff(F[2], x)-(diff(F[1], y))];

[2, sin(z), 2*x]


#Matrise for kryssprodukt CURL=GradientF X F="[[[(&PartialD;)/(&PartialD; x) ,(&PartialD;)/(&PartialD; y) ,(&PartialD;)/(&PartialD; z) ],[F[1],F[2],F[3]]]]:"NULL

intergrand := DotProduct(CurlF, N); 1; CurlF; 1; N



[2, sin(z), 2*x]


[2, 0, 1]


int(intergrand, [y = -sqrt(-x^2-2*x) .. sqrt(-x^2-2*x), x = -2 .. 0])



int((2*(r*cos(t)+2))*r, [r = 0 .. -2*cos(t), t = (1/2)*Pi .. 3*Pi*(1/2)]);




a := Int(intergrand, [y = -sqrt(-x^2-2*x) .. sqrt(-x^2-2*x), x = -2 .. 0])

Int(4+2*x, [y = -(-x^2-2*x)^(1/2) .. (-x^2-2*x)^(1/2), x = -2 .. 0])


b := ChangeOfVariables(a, [cartesian[x, y], polar[r, theta]])

Student:-MultivariateCalculus:-ChangeOfVariablesObject(Int(Int(2*(r*cos(theta)+2)*r, r = (y = -(-x^2-2*x)^(1/2)) .. (y = (-x^2-2*x)^(1/2))), theta = (x = -2) .. (x = 0)), s)




Download stokes_surface_riktig.mw



Thanks, How can I use this to find the bounds for R if I was to set up a work integral. To find the area I can use r=0..2 but for a work integral I would need to express r differently.

Thanks to both of you :)

I applied orientation=[0,0,0] that gave me the top view as well


Thank you :)

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