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Hi,

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:
            
with(plots):
with(plottools):
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);

p3[3]:
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.

Thanks.



 

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.

Thanks

Hi,

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? http://math.stackexchange.com/questions/407310/how-can-i-find-the-surface-area-of-a-normal-chicken-eggPlaying 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

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?

Question.doc

Can you make geometric figures that hold a particular area within a given perimeter of 12 matches

My attempt was to start at origin 0,0 and move up 1 unit (a,b) and then 1 unit up or to the right (c,d) etc ... until you enclose the required area (defined by linear spline) s.t. perimeter =12. but it doesn't really work

Hi. I'm very new to maple, and got stuck on a task on my assignment. I'm supposed to find the area of the segment (gray area) with the diameter D as a function of h. Any help and pointers is appreciated, thanks. :)

 

Circle

Hi folks,

I'm trying to find the x value of my data set that corresponds to 50% of area under the curve. I've used curvefitting and splines to generate a piecewise function, and am able to find the area under the curve between my two approximated zeros (as this is the only relevant region). I would greatly appreciate any help, as I've been pulling my hair out for a couple hours now!

 

Cheers!

I'm trying to plot and fill the area between these two curves from x[0,450]

line1:=20-0.05*x

line2:=2+0.0002*x^2

 

I can do it separately but when I do them together, it gives me extra area that isn't shared between the two curves.

Also, whenever I try to plot them together I get the following error message:

Error, (in plot) unexpected option: 2+0.2e-3*x^2

 

Any help will be greatly appreciated!

I would like to draw some areas between curves. I tried to use ?plots/implicitplot but I often need to use gridrefine option which consumes so much memory. When I want to draw e.g. the area between f and g:

f := x-> piecewise(x < 5, 2, x-3):
g := x-> piecewise(x < 4, x, 4):

I can do:

p:=plot({f, g}, color...

I have the fractcal of a snowflake

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