help me for my project

this is the question

r^2=2cosθ ,θ=-π/6   and θ=π/4

how can i solve it ?

and i want describe the method in detail and compare with other methods if they exist. and give simulation results using software

also,state the weaknesses of the work and possible improvement

please help me :(

[Walter Roberson]

Please read the guide to tags and retag this question. See http://www.mathworks.co.uk/matlabcentral/answers/43073-a-guide-to-tags

[Walter Roberson]

How can theta be two different values at the same time?

What is there to do other than sqrt(2 * cos(theta)) ?

[Fahad almulhim]

that's what i faced , it's confused me ! but do you think can i solve it by integration ?

[Walter Roberson]

Is there an integral sign that got missed when you posted?

[Fahad almulhim]

No, there is not.

[Walter Roberson]

Please confirm whether it is intended to be a polar plot.

[Fahad almulhim]

No,it isn't intended to be a polar plot.

[Walter Roberson]

The question is undefined if it is not a polar plot. The angles would not have a reference point, and "r" has no meaning in an x-y graph.

[Cedric]

Could be y(x=theta) = r(theta) = sqrt(2*cos(theta)), with an orthonormal basis.

Fahad, are you able to plot the function that you want to integrate? If not, are you at least able to compute its values on a vector of thetas in the relevant range?

Answer 1

[Jan]

theta1 = -pi / 6;
r1 = sqrt(2 * cos(theta1));

theta2 = pi / 4;
r2 = sqrt(2 * cos(theta2));

It is hard to describe such a simple method in detail. It is not clear, with other methods this solution should be compared. It is also not clear, how this could be "simulated". And I do not see either a weakness nor a possible improvement.

[Fahad almulhim]

oky, how i can find the area between the graph of r^2=2cosθ the origin , and between the rays θ=-π/6 and θ=π/4 . when we will ussing by plotting ?

[Walter Roberson]

That is a quite different question than your original.

Is "r" a fixed value, or is the expression intended to be a polar plot ?

[Fahad almulhim]

i know , let tell you the original question... find the area between the graph of r^2=2cosθ the origin , and between the rays θ=-π/6 and θ=π/4 and original. ussing by plotting

[Fahad almulhim]

and can i get conclusion for this question to my project?

[Jan]

It wastes the time of yourself, of the readers and the ones who answer, when you do not post the original question initially. Now this thread has an accepted answer, such that less readers will care about it, and the actual question is hidden in a comment, where readers will not find it.

Please post, what you have tried so far.

[Walter Roberson]

It did get reposted as a new Question, but I closed it as a duplicate as a solution had already been given here by the time the duplicate was created.

Answer 2

[Walter Roberson]

Use polar integration; the formula is at http://en.wikibooks.org/wiki/Calculus/Polar_Integration

The integration will be pretty trivial. You can do it by hand or you can use the symbolic toolbox.

And then to verify, see http://www.mathworks.com/discovery/integral.html

[Fahad almulhim]

my question say , find the area between the graph of r^2=2cosθ , the origin,and between the rays θ=-π/6 and θ=π/4. Use visual aids by plotting all graphs and the desired region.

[Walter Roberson]

The above answers your question about finding the area. Use polar() to do a polar plot.