SLD
Contributor
Integrate the function x^2+3y over the circle of radius 1 centered at y=1, x=0.
Integrate the function x^2+3y over the circle of radius 1 centered at y=1, x=0.
Integrate the function x^2+3y over the circle of radius 1 centered at y=1, x=0.
=7*Pi
It's also helpful to sanity-check this sort of problem by bounding it. Integration is linear, so it's the same problem as these two:... = (13/4)*pi
I checked my work with Mathematica.
Wait, it's 2D integral?
over a circle means a line.
Regardless, since I integrated over a circle (1D) and it is Pi*(6*r + r^3). all I need to do is to integrate over radius [0,1]
which is Pi*(3+1./4)
You should have said "inside a circle" instead of "over a circle", or "over a disk"