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Hello, i am doing some schmidt-analysis on a stirling engine, but my question is rather simple. I have the measured presure P at a given time T as a 229x2 matrix, i also have a function, V__total(T), for the total volume of the engine at a given time T. 

I then go on to create a pointplot PV, which is rougly the shape of a potato. I now want to find the area enclosed by this point plot, is there any way? 

I do something like:

> DATA:=ImportMatrix(filepath,skiplines=1); %import data from .txt file, skip header line. 

>P:=DeleteColumn(DATA,1); %Isolate presure column

>T:=DeleteColumn(DATA,2); %Isolate time coliumn

>V:=V__total~(T); %Generate volume vector as a function of time T


Is there any wat to finde the area enclosed by the curve/ Int(P,V)


I would like to draw a spectrum of plots. For example sin(mx) when m ∈ [1,2] and x ∈ [0,2] that is a spectrum of sinous plots. Obviously, if I write

plot(sin(m*x), m = 1 .. 2, x = 0 .. 2)

I will receive the error

Warning, expecting only range variable m in expression sin(m*x) to be plotted but found name x

I expect that there exist an alternative plot command in Maple, able to do that. I checked some members of Plots family that their name conducted me to check them. Moreover, I searched the web: but the most common retrieved results are related to drawing the area between two plots or under one plot.

A real example of what I am looking for is (not exaclty but something like) the below picure

A plot of Gaussian plots

Could you please let me know the alternative plot command, satisfying my requirement?

I related two Text Area Components by using the codes "Do(%text_beta_radian=evalf(%text_beta_degress/180*Pi))".

I wanna know what codes can refresh the display of the text area component with name%text_beta_radian

after I change the input value of that%text_beta_degress?

and plot  function I? This I is the area which I wrote at the paper.

Could you give me the code which can be used to solve the ODE by numerical method and plot I with respect to t?

I think I have write down everything clearly but if you feel confused please ask me.

I am eager to know the code. Thanks very much!

Hi to all

I had solve a set of ODEs using rkf45 method and I had plot it's Diagram now I need to have area under my diagram.what should I do? what is code?


The downloaded worksheet below displays 3 points on the unit sphere which define a solid angle with a triangular face. The sides of the solid angle's are red arcs on the surface of the sphere and red radii which outline the planar sides within the sphere.

Three questions:

1. Is there a way to make the surfaces of the solid triangle more apparent by filling them with color?

2. Is there a way to calculate the area of the face on the surface of the sphere?

3. Is there a way to calculate the volume of the solid triangle?


Download Mechanics;


I have a small problem. I want to findout area under a curve. I got the plot from solving a partial differential equation. I want to find out area under the curve with out using interpolation. Are there any methods to find this.


here i enclose the method i have done.

Es := 0.117108e12:
Ef := 0.78125e11:
l := 0.150e-6:
s := 0.500000e-3:
f := 0.5898334197e-6:
o := 0.9e-5:
d := 0.10e-17:
cb := 0.1e7/(19.9):
c := l*f/(d*cb):

PDE := diff(u(x, t), t)-(diff(u(x, t), x, x)) = 0:
ys := -0.4245333333e-1:
IBC1 := {u(x, 0) = 0, (D[1](u))(0, t) = 0, (D[1](u))(1, t) = c}:
S1 := pdsolve(PDE, IBC1, numeric, time = t, timestep = 0.1e-2);

p2 := S1:-plot(t = .2525);

p3 := getdata(p2);

co:=CurveFitting[PolynomialInterpolation](p3[3], x):
Area := int(co, x = x[1] .. x[2]):

So this is the procedure i used to find out, but can there be any other procedure to findout area directly from hte solution of PDE.



Not sure exactly how i could achieve this but:

how do i determine the value of k for which the graphs p(x) = x^2+2x+3 and q(x) = k+5x-7x^2 enclose an area of exactly 36?

I have to do it in maple and using i guess area under the curve.



I'm pretty new into MAPLE andI'm trying to get into it.

I have these four equations:

eq1:=1.6*10^(-7)*R*sin(t)-4.4*10^(-14)*R^2*cos(t)*sin(t)-1.6*10^(-14)*R^2*cos(t)^2+4.2*10^(-14)*R^2-1.3+2.1*10^(-9)*R*cos(t)=0; eq2 :=8.3*10^(-8)*R*sin(t)-1.2*10^(-13)*R^2*cos(t)*sin(t)-2.9*10^(-44)*R^2*cos(t)^2+7.1*10^(-14)*R^2-1.3+8.3*10^(-8)*R*cos(t)=0; eq3 :=8.3*10^(-8)*R*sin(t)-1.2*10^(-13)*R^2*cos(t)*sin(t)-2.2*10^(-44)*R^2*cos(t)^2+7.1*10^(-14)*R^2-1.3+8.3*10^(-8)*R*cos(t)=0; eq4 :=2.1*10^(-9)*R*sin(t)-4.4*10^(-14)*R^2*cos(t)*sin(t)+1.6*10^(-14)*R^2*cos(t)^2+2.6*10^(-14)*R^2-1.3+1.6*10^(-7)*R*cos(t)=0;

Hi Maple friends.

If I have y=-x^2+9 and y=x+3, how can I find the area that is bounded by the functions? 

Easy enough to calculate by hand, but I need to use Maple to check my answers. Most preferably I'd like to use the Maple context menu to get the result, but any solution will be appreciated.

Thanks in advance.

Hi Maple friends.

As per usual, I am unable to find the solution in Maple Help. Can someone please advice me on how to shade an area under a curve, preferably using the context menu, since that is what I prefer to use at this stage.

Thanks in advance.

How can I find the surface area of a normal chicken egg? replies to the topic in Mathematics Stack Exchange:_SurfaceAreaChickenE.zipGracias to Acer and Carl Love for your answers.

Given a figure in the plane bounded by the non-selfintersecting piecewise smooth curve. Each segment in the border defined by the list in the following format (variable names  in expressions can be arbitrary):

1) If this segment is given by an explicit equation, then  [f(x), x=x1..x2)]

2) If it is given in polar coordinates, then  [f(phi), phi=phi1..phi2, polar] , phi is polar angle

3) If the segment is given parametrically, then  [[f(t), g(t)], t=t1..t2]

4) If several consecutive segments or entire border is a broken line, then it is sufficient to set vertices the broken line [ [x1,y1], [x2,y2], .., [xn,yn]]


The first procedure symbolically finds perimeter of the figure. Global variable  Q  saves the lengths of all segments.

Perimeter := proc (L) #  L is the list of all segments of the border

local i, var, var1, var2, e, e1, e2, P;

global Q;

for i to nops(L) do if type(L[i], listlist(algebraic)) then P[i] := seq(simplify(sqrt((L[i, j, 1]-L[i, j+1, 1])^2+(L[i, j, 2]-L[i, j+1, 2])^2)), j = 1 .. nops(L[i])-1) else

var := lhs(L[i, 2]); var1 := min(lhs(rhs(L[i, 2])), rhs(rhs(L[i, 2]))); var2 := max(lhs(rhs(L[i, 2])), rhs(rhs(L[i, 2])));

if type(L[i, 1], algebraic) then e := L[i, 1]; if nops(L[i]) = 3 then P[i] := simplify(int(sqrt(e^2+(diff(e, var))^2), var = var1 .. var2)) else

P[i] := simplify(int(sqrt(1+(diff(e, var))^2), var = var1 .. var2)) end if else

e1 := L[i, 1, 1]; e2 := L[i, 1, 2]; P[i] := abs(simplify(int(sqrt((diff(e1, var))^2+(diff(e2, var))^2), var = var1 .. var2))) end if end if end do;

Q := [seq(P[i], i = 1 .. nops(L))];

add(Q[i], i = 1 .. nops(Q));

end proc:


The second procedure symbolically finds the area of the figure. For correct work of the procedure, all the segments in the list L  of border must pass sequentially in clockwise or counter-clockwise direction.

Area := proc (L)

local i, var, e, e1, e2, P;

for i to nops(L) do

if type(L[i], listlist(algebraic)) then P[i] := (1/2)*add(L[i, j, 1]*L[i, j+1, 2]-L[i, j, 2]*L[i, j+1, 1], j = 1 .. nops(L[i])-1) else

var := lhs(L[i, 2]);

if type(L[i, 1], algebraic) then e := L[i, 1];

if nops(L[i]) = 3 then P[i] := (1/2)*(int(e^2, L[i, 2])) else

P[i] := (1/2)*simplify(int(var*(diff(e, var))-e, L[i, 2])) end if else

e1 := L[i, 1, 1]; e2 := L[i, 1, 2]; P[i] := (1/2)*simplify(int(e1*(diff(e2, var))-e2*(diff(e1, var)), L[i, 2])) end if end if

end do;

abs(add(P[i], i = 1 .. nops(L)));

end proc:


The third procedure shows this figure. To paint the interior of the boundary polyline approximation is used. Required parameters: L - a list of all segments of the border and C - the color of the interior of the figure in the format color = color of the figure. Optional parameters: N - the number of parts for the approximation of each segment (default N = 100) and Boundary is defined by a list for special design of the figure's border (the default border is drawed by a thin black line). The border of the figure can be drawn separately without filling the interior by the global variable Border.

Picture := proc (L, C, N::posint := 100, Boundary::list := [linestyle = 1])

local i, var, var1, var2, e, e1, e2, P, Q, h;

global Border;

for i to nops(L) do

if type(L[i], listlist(algebraic)) then P[i] := op(L[i]) else

var := lhs(L[i, 2]); var1 := lhs(rhs(L[i, 2])); var2 := rhs(rhs(L[i, 2])); h := (var2-var1)/N;

if type(L[i, 1], algebraic) then e := L[i, 1];

if nops(L[i]) = 3 then P[i] := seq(subs(var = var1+h*i, [e*cos(var), e*sin(var)]), i = 0 .. N) else

P[i] := seq([var1+h*i, subs(var = var1+h*i, e)], i = 0 .. N) end if else

e1 := L[i, 1, 1]; e2 := L[i, 1, 2]; P[i] := seq(subs(var = var1+h*i, [e1, e2]), i = 0 .. N) end if end if

end do;

Q := [seq(P[i], i = 1 .. nops(L))];

Border := plottools[curve]([op(Q), Q[1]], op(Boundary));

[plottools[polygon](Q, C), Border];

end proc:


Examples of works:

Example 1.

L := [[sqrt(-x), x = -1 .. 0], [2*cos(t), t = -(1/2)*Pi .. (1/4)*Pi, polar], [[1, 1], [1/2, 0], [0, 3/2]], [[-1+cos(t), 3/2+(1/2)*sin(t)], t = 0 .. -(1/2)*Pi]];

Perimeter(L); Q; evalf(`%%`); evalf(`%%`); Area(L); 

plots[display](Picture(L, color = grey, [color = "DarkGreen", thickness = 4]), scaling = constrained);

plots[display](Border, scaling = constrained);

Example 2.

The easiest way to use this  procedures for polygons.

 L := [[[3, -1], [-2, 2], [5, 6], [2, 3/2], [3, -1]]];

Perimeter(L), Q;


plots[display](Picture(L, color = pink, [color = red, thickness = 3]));



Example 3 (more complicated )

3 circles on the plane C1, C2 and C3 defined by the parametric equations  of their borders. We want to find the perimeter, area, and paint the figure  C3 minus (C1 union C2) . For details see attached file. 

C1 := {x = -sqrt(7)+4*cos(t), y = 4*sin(t)};

C2 := {x = 3*cos(s), y = 3+3*sin(s)};

C3 := {x = 4+5*cos(u), y = 5*sin(u)};

L := [[[-sqrt(7)+4*cos(t), 4*sin(t)], t = -arccos((1/4)*(7+4*sqrt(7))/(sqrt(7)+4)) .. -arctan((3*(-23+sqrt(7)*sqrt(55)))/(23*sqrt(7)+9*sqrt(55)))], [[3*cos(s), 3+3*sin(s)], s = -arctan((1/3)*(9+sqrt(7)*sqrt(55))/(-sqrt(7)+sqrt(55))) .. arctan((1/3)*(-9+4*sqrt(91))/(4+sqrt(91)))], [[4+5*cos(u), 5*sin(u)], u = arctan((3*(41+4*sqrt(91)))/(-164+9*sqrt(91)))+Pi .. arctan(3/4)-Pi]];

Perimeter(L), Q; evalf(%);

Area(L); evalf(%)

 A := plot([[rhs(C1[1]), rhs(C1[2]), t = 0 .. 2*Pi], [rhs(C2[1]), rhs(C2[2]), s = 0 .. 2*Pi], [rhs(C3[1]), rhs(C3[2]), u = 0 .. 2*Pi]], color = black);

B := Picture(L, color = green, [color = black, thickness = 4]);

plots[display](A, B, scaling = constrained);

More examples and all codes see in attached file

I was thinking about the area problem, yet again, and found myself asking the question: why must we go through such elaborate means to get Maple to generate a plot of the region between two (or more curves)? I use the word elaborate to describe any process that would might become overwhelming, for, say a student, to go through to accomplish a task. Anyone with the most basic of backgrounds can understand the area problem, but yet, such an individual might not find it a trivial...

How to color only certain bounded/intersection area of a graph in Maple?

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