I need help in finding the area of the shaded region in terms of w,a,b

# Getting complex value for real integration

2 views (last 30 days)

Show older comments

Mithun Kumar Munaganuri
on 31 Mar 2022

Commented: Mithun Kumar Munaganuri
on 5 Apr 2022

### Accepted Answer

David Goodmanson
on 4 Apr 2022

Edited: David Goodmanson
on 4 Apr 2022

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

### More Answers (0)

### See Also

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!