Above is the graph I have obtained from the equation.
Hi there so I've just been introduced to using polar co-ordinates to find double integrals and Im having a hard time working out how to tackle questions step by step.
Im currently looking at the question:
$I_{6}=\int\int_{Q}x/\pi (x^{2}+y^{2}) dxdy$ with $Q=\{(x,y)|(x-3/2)^2 +y^2 \leq9/4, y\geq0\}$
And Im struggling to make a start to this as I dont know how to obtain the limits of the double integral from $Q$ .
All I know is that I should eventually obtain the answer 3/4.
Any help would be greately appreciated.
Update:
So I now have the double integral
$\int_{0}^{\pi/2}\int_{0}^{3cos\theta}(rcos\theta/\pi)*((rcos\theta)^{2}+(rsin\theta)^{2}) rdrd\theta$
On your specific question about how the equation in polar form becomes $r = 3 \cos \theta$,
In polar form, $x = r \cos \theta, y = r \sin \theta$
Now your curve is $(x-3/2)^2 +y^2 = 9/4 \implies x^2 + y^2 = 3x$
$r^2 cos^2 \theta + r^2 \sin^2 \theta = 3r \cos \theta$
$r^2 = 3r \cos \theta \implies r = 3 \cos \theta$
Here is a diagram showing how $r$ changes with $\theta$ with $x$-axis.