Finding the double integral using polar coordinates.

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

$$\int_{0}^{\pi/2}\int_{0}^{3cos\theta}(rcos\theta/\pi)*((rcos\theta)^{2}+(rsin\theta)^{2}) rdrd\theta$$

• Have you written down the equation of the region in polar form, which is a circle with center at $(3/2, 0)$? Commented Nov 30, 2020 at 13:04
• No I haven't . I dont actually know how to do that. Could you explain any further on how to do this please.
– xyz
Commented Nov 30, 2020 at 13:08
• Equation of your region $Q$ will be $r = 3 \cos \theta$ in polar form $(0 \leq \theta \leq \pi/2)$. Commented Nov 30, 2020 at 13:11
• Thanks. So should my equation look like $I_{6}=\int\int_{Q} (rcos\theta)/(\pi) ((rcos\theta)^2+(rsin\theta)^2) drd\theta$ ?
– xyz
Commented Nov 30, 2020 at 13:19
• Okay thank you and what would my limits be for r dr. I think from your comments the limit for $d\theta$ will be 0 and $\pi/2$
– xyz
Commented Nov 30, 2020 at 13:31

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.

• That does not seem right. $r^2 cos^2\theta + r^2\sin^2\theta$ will simply give you $r^2$. So simplify and then integrate. Commented Nov 30, 2020 at 14:18
• what about the $r$ from the Jacobian? :) Always remember in polar $dx \, dy$ is $r \, dr \, d\theta$. Commented Nov 30, 2020 at 14:29
• You can also sketch and try to see $dA$ in terms of $r, \theta$. It will be clear why $dA = r \, dr \, d\theta$. Commented Nov 30, 2020 at 14:32
• Circle is drawn correct but the radius should be drawn from the origin. That is how radius is a function of $\theta$ otherwise radius will be independent of $\theta$. $\cos 0^0 = 1, r = 3$, $\cos \pi/2 = 0, r = 0$... Commented Nov 30, 2020 at 16:35
• It depends on the center of the circle. If it is on $(0,0)$ then distance of points on the circle from the origin always continues to be the same so to cover the circle, you have to go form $0$ to $2\pi$. If the center is on $y$-axis away from origin instead of on x-axis then it will be different. Commented Dec 1, 2020 at 10:05