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How to display the coordinate set of a g...

Oliveira 180
November 06 2022
0 0

It's what I was needing. I am very grateful for your attention mmcdara.

Regards,

Oliveira

How to display the coordinate set of a g...

Oliveira 180
November 05 2022
0 0

Sorry to bother you again acer. I didn't get the same result with the example below:

with(plots):

s1 := x^2 + y^2 + z^2 = 1;
s2 := x^2 - y^2 = z*(z - 1);
pon1 := intersectplot(surface(s1, x = -1 .. 1, y = -1 .. 1, z = -1 .. 1), surface(s2, x = -1 .. 1, y = -1 .. 1, z = -1 .. 1), axes = box, thickness = 4, orientation = [70, 40]);

temp1:=plottools:-transform((x,y,z)->[x,y,z])(pon1):
convert(plottools:-getdata(temp1)[3],listlist);

How to display the coordinate set of a g...

Oliveira 180
November 05 2022
0 0

This what I was needing. I appreciate your attention acer.

Regards

Oliveira

How to display the coordinate set of a g...

Oliveira 180
November 05 2022
0 0

Applying the procedure provided by Kotonum 19009, the coordinates do not appear in the form [x,y,z]. I would like the points to appear in the form of vectors or lists.

Triple integral with assumptions...

Oliveira 180
October 01 2022
0 0

@Axel Vogt 

An attached document follows.
 

with(VectorCalculus)

> 

``

This problem I took from a book, which asks to determine the magnetic field H inside a solid spherical cone.

 

             

 

The author solved the question using Ampere's circuit law and found the following expression:

 

                    

 

         

 

 

I tried to solve the issue using integration through FF expression:

  

                   

 

NULL

 

NULL

The coordinates of the infinitesimal volume element dV is:

 

rv = {p*cos(alpha), p*sin(alpha)*cos(phi), p*sin(alpha)*sin(phi)}

 

The coordinates of point P on the cone are:

 

rp = {0, r*cos(theta), r*sin(theta)}

 

Integration will be performed for:

 

" 0<p<1 0<alpha<Pi/(4) 0<phi<2 Pi d(H)=((J)/(4 Pi r^(2))*(r)[u])dv H=(&int;)[0]^(2Pi)(&int;)[0]^(Pi/(4))(&int;)[0]^(1)((J)/(4 Pi r^(2))*(r)[u])p^(2)sin(alpha) &DifferentialD;p &DifferentialD;alpha &DifferentialD;phi"
NULL

NULL

> 

rR := `<,>`(p*sin(alpha)*cos(phi), p*sin(alpha)*sin(phi), p*cos(alpha))

Vector(3, {(1) = p*sin(alpha)*cos(phi), (2) = p*sin(alpha)*sin(phi), (3) = p*cos(alpha)})``

 (1)
> 

dA := `assuming`([simplify(sqrt(`&x`(diff(rR, phi), diff(rR, alpha)).`&x`(diff(rR, phi), diff(rR, alpha))))], [p > 0, alpha > 0, alpha < (1/4)*Pi])
NULL

p^2*sin(alpha)

 (2)
> 

AA := int(int(dA, alpha = 0 .. (1/4)*Pi), phi = 0 .. 2*Pi)

2*p^2*(1-(1/2)*2^(1/2))*Pi

 (3)
> 

rv := `<,>`(p*sin(alpha)*cos(phi), p*sin(alpha)*sin(phi), p*cos(alpha))

Vector(3, {(1) = p*sin(alpha)*cos(phi), (2) = p*sin(alpha)*sin(phi), (3) = p*cos(alpha)})

 (4)
> 

rp := `<,>`(r*sin(theta), 0, r*cos(theta))

Vector(3, {(1) = r*sin(theta), (2) = 0, (3) = r*cos(theta)})

 (5)
> 

rr := rp-rv

Vector(3, {(1) = r*sin(theta)-p*sin(alpha)*cos(phi), (2) = -p*sin(alpha)*sin(phi), (3) = r*cos(theta)-p*cos(alpha)})

 (6)
> 

ru := `assuming`([simplify(rr/sqrt(rr.rr))], [p > 0, alpha > 0, alpha < (1/4)*Pi])

Vector(3, {(1) = (r*sin(theta)-p*sin(alpha)*cos(phi))/sqrt(-2*sin(theta)*sin(alpha)*cos(phi)*p*r-2*cos(alpha)*cos(theta)*p*r+p^2+r^2), (2) = -p*sin(alpha)*sin(phi)/sqrt(-2*sin(theta)*sin(alpha)*cos(phi)*p*r-2*cos(alpha)*cos(theta)*p*r+p^2+r^2), (3) = (r*cos(theta)-p*cos(alpha))/sqrt(-2*sin(theta)*sin(alpha)*cos(phi)*p*r-2*cos(alpha)*cos(theta)*p*r+p^2+r^2)})

 (7)
> 

rJ := subs(J = i/AA, Typesetting[delayDotProduct](J, `<,>`(sin(alpha)*cos(phi), sin(alpha)*sin(phi), cos(alpha)), true))

Vector(3, {(1) = (1/2)*i*sin(alpha)*cos(phi)/(p^2*(1-(1/2)*sqrt(2))*Pi), (2) = (1/2)*i*sin(alpha)*sin(phi)/(p^2*(1-(1/2)*sqrt(2))*Pi), (3) = (1/2)*i*cos(alpha)/(p^2*(1-(1/2)*sqrt(2))*Pi)})

 (8)
> 

dH := `assuming`([simplify(subs(dv = p^2*sin(alpha), `&x`(rJ/(4*Pi(rr.rr)), ru)*dv))], [p > 0, alpha > 0, alpha < (1/4)*Pi])

Vector(3, {(1) = -(1/4)*sin(alpha)^2*i*sin(phi)*r*cos(theta)/(sqrt(-2*sin(theta)*sin(alpha)*cos(phi)*p*r-2*cos(alpha)*cos(theta)*p*r+p^2+r^2)*Pi(-2*sin(theta)*sin(alpha)*cos(phi)*p*r-2*cos(alpha)*cos(theta)*p*r+p^2+r^2)*(-2+sqrt(2))*Pi), (2) = -(1/4)*sin(alpha)*i*r*(-sin(alpha)*cos(phi)*cos(theta)+sin(theta)*cos(alpha))/(sqrt(-2*sin(theta)*sin(alpha)*cos(phi)*p*r-2*cos(alpha)*cos(theta)*p*r+p^2+r^2)*Pi(-2*sin(theta)*sin(alpha)*cos(phi)*p*r-2*cos(alpha)*cos(theta)*p*r+p^2+r^2)*(-2+sqrt(2))*Pi), (3) = (1/4)*sin(alpha)^2*i*sin(phi)*r*sin(theta)/(sqrt(-2*sin(theta)*sin(alpha)*cos(phi)*p*r-2*cos(alpha)*cos(theta)*p*r+p^2+r^2)*Pi(-2*sin(theta)*sin(alpha)*cos(phi)*p*r-2*cos(alpha)*cos(theta)*p*r+p^2+r^2)*(-2+sqrt(2))*Pi)})

 (9)
> 

dh1 := `assuming`([simplify(int(dH, p = 0 .. 1))], [alpha > 0, alpha < (1/4)*Pi, r > 0, r < 1, phi > 0, phi < 2*Pi, theta > 0, theta < (1/4)*Pi])

Vector(3, {(1) = -(1/4)*sin(alpha)^2*i*sin(phi)*r*cos(theta)*(int(1/(sqrt(-2*sin(theta)*sin(alpha)*cos(phi)*p*r-2*cos(alpha)*cos(theta)*p*r+p^2+r^2)*Pi(-2*sin(theta)*sin(alpha)*cos(phi)*p*r-2*cos(alpha)*cos(theta)*p*r+p^2+r^2)), p = 0 .. 1))/((-2+sqrt(2))*Pi), (2) = -(1/4)*sin(alpha)*i*r*(-sin(alpha)*cos(phi)*cos(theta)+sin(theta)*cos(alpha))*(int(1/(sqrt(-2*sin(theta)*sin(alpha)*cos(phi)*p*r-2*cos(alpha)*cos(theta)*p*r+p^2+r^2)*Pi(-2*sin(theta)*sin(alpha)*cos(phi)*p*r-2*cos(alpha)*cos(theta)*p*r+p^2+r^2)), p = 0 .. 1))/((-2+sqrt(2))*Pi), (3) = (1/4)*sin(alpha)^2*i*sin(phi)*r*sin(theta)*(int(1/(sqrt(-2*sin(theta)*sin(alpha)*cos(phi)*p*r-2*cos(alpha)*cos(theta)*p*r+p^2+r^2)*Pi(-2*sin(theta)*sin(alpha)*cos(phi)*p*r-2*cos(alpha)*cos(theta)*p*r+p^2+r^2)), p = 0 .. 1))/((-2+sqrt(2))*Pi)})

 (10)
> 

``

Note that the result of the first integration is already quite complicated and I couldn't go any further.

``


 

Download Example1.mw

Triple integral with assumptions...

Oliveira 180
October 01 2022
0 0

@acer 

Thanks for the fix acer.

Triple integral with assumptions...

Oliveira 180
October 01 2022
0 0

Thank you for your attention mmcdara. In fact formal integration is complicated, so I'm going to try numerical integration.
Regards,
Oliveira

Re-Try...

Oliveira 180
September 02 2019
0 0

Thanks again for the tip.

Olveira

Re-Try...

Oliveira 180
September 02 2019
0 0
Thank you for your attention Mariusz. I have difficulty calculating the gradient with this numerical procedure.

 

eval(-Gradient(uval(r, t), 'cylindrical'[r, theta, z]), [r = 2, t = 4])

Re-Examples...

Oliveira 180
September 01 2019
0 0

Thanks for the reply and attention.

Oliveira 

Re-The trouble with temperature...

Oliveira 180
August 29 2019
0 0

Thanks for the tips, because I'm not familiar with some Maple functions yet.

Re-convert, temperature...

Oliveira 180
August 29 2019
0 0

Thanks for the tips, because I'm not familiar with some Maple functions yet.

Re-useint...

Oliveira 180
August 12 2019
0 0
I am grateful for the information.

Re-little help...

Oliveira 180
August 12 2019
0 0

Thanks for the tip.

Re-Fixed; Physics Updates v.416...

Oliveira 180
August 11 2019
0 0
I installed the new version of the Physics package and problem was solved.
Thanks again for your attention.
Oliveira
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