multiple integration §O (x^2+y^2)^(1/2) dxdy?
O={(x,y)€R^2: (x^2+y^2-4x)<0}
it's number p)
http://calvino.polito.it/~nicola/analisi-II/Eserci...
I cannot understand the domain decomposition from
O={(x,y)€R^2: (x^2+y^2-4x)<0}
to O'={(x,y)€R^2: 0<r<4costh, -pi/2<th<pi/2}
can anyone show me step by step which passages are made here to obtain the second domain? thanks
Update:thank you,
I don't understand
why is Q between -pi/2 and +pi/2?
My idea is that since in polar coordinates
x=rcosQ
y=rsinQ
and x^2+y^2=r^2
if I have x^2+y^2<4x
I can change it into r^2<4rcosQ
that is r<4cosQ
is it correct?
then I don't understand the rest, that is why I have to modify the domain to 0<r<4cosQ
why has r to be >0
and why I have to consider only the section with Q between -pi/2 and pi/2.
what is the operation behind it?
thanks
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Verified answer
Complete the square:
x^2 + y^2 - 4x = (x - 2)^2 + y^2 - 4
So the inequality given is equivalent to
(x - 2)^2 + y^2 < 4
which is the interior of the circle of radius 2, centered at (2,0). (So the circle is tangent to the y-axis.) In polar coordinates, the equation of that circle is r = 4 cos Q (where Q is the angle). Now you can easily see that the interior of the circle in polar coordinates is given by the inequality 0 < r < 4 cos Q, where Q is between -pi/2 and +pi/2 (negative y-axis to positive y-axis).