Getting complex value for real integration
4 ビュー (過去 30 日間)
古いコメントを表示
Mithun Kumar Munaganuri
2022 年 3 月 31 日
コメント済み: Mithun Kumar Munaganuri
2022 年 4 月 5 日
I'm getting complex value in real integration while trying to find area under ellipse as shown in below figure. Please advise
9 件のコメント
採用された回答
David Goodmanson
2022 年 4 月 4 日
編集済み: David Goodmanson
2022 年 4 月 4 日
Hi Mithun,
One of the more straightforward methods is to forget about symbolics and just write down the solution for y. The ellipse has semimajor axis a along x, semiminor axis b along y, and is centered at
( w/2, -(b/(2*a))*sqrt(4*a^2-w^2) )
so that the ellipse passes through both the origin and (w,0). w has to be less than 2*a, the major axis. Solve the ellipse equation for y,
( y + (b/(2*a))*sqrt(4*a^2-w^2) )^2/b^2 = 1 - (x-w/2).^2/a^2
take the sqrt of both sides**, rearrange
y = -(b/(2*a))*sqrt(4*a^2-w^2) +- b*sqrt((1-(x-w/2).^2/a^2)).
** the sqrt of the left hand side is taken as positive. The sqrt of the right hand side can be either sign, but the positive sqrt is chosen since from the figure, y>=0. So
a = 3; b = 2; w = 5;
fun = @(x) -(b/(2*a))*sqrt(4*a^2-w^2) + b*sqrt((1-(x-w/2).^2/a^2));
integral(fun,0,w)
ans = 3.1468
3 件のコメント
David Goodmanson
2022 年 4 月 4 日
Hi Mithun,
you're most welcome. One thing I forgot to mention is that when you post a question it's not best practice to post images of code. It's much better to copy in the code as text. That way, people looking to assist can just copy the code and run it, rather than having to type it in by hand.
その他の回答 (0 件)
参考
カテゴリ
Help Center および File Exchange で Lighting, Transparency, and Shading についてさらに検索
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!