Vtyjkyuk

I have to calculate the next integral in polar coordinates but it confuses me to calculate the extremes of integration. The condition $x> 1$ confuses me.

$iint_D 6x dx dy$

$D=(x,y)in mathbbR^2:ygeq 0,xgeq 1, x^2+y^2 leq 4$

Thanks!

The domain is the region bounded by a circle centered at the origin with radius $2$, the line $y=0$ and the line $x=1$ with the condition $xge 1$ that is

therefore

• $0le theta le fracpi3$


• $r_min(theta)=frac 1 cos thetale r le 2$

therefore the set up becomes

$$iint_D 6x ,dx dy=int_0^fracpi3, dthetaint_frac 1 cos theta^26r^2cos theta ,dr$$

Hint:

The equation of a vertical line $x=k$ in polar coordinates is simply $rcos theta=k$, so the bounds for $r$ in the integral in polar coordinates are
$$int_tfrac1costheta^2 ...,mathrm dr$$

in polar coordinates $x=rcos(theta)$ and $y=rsin(theta)$

also $x^2+y^2le 4$

put $x=rcos(theta)$ and $y=rsin(theta)$ in $x^2+y^2le 4$
gives $rle2$

given $yge 0$ and $x ge 1$ thus $y^2le 3$

means $rsin(theta)le sqrt3$

$2sin(theta)le sqrt3$

$sin(theta)le sqrt3/2$
thus $theta le pi/3$

$$iint_D 6x ,dx dy=int_0^fracpi3, cos theta dthetaint_frac 1 cos theta^26r^2,dr$$