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enter image description here

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

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

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

1 Answer 1

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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.

enter image description here

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    $\begingroup$ 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. $\endgroup$
    – Math Lover
    Commented Nov 30, 2020 at 14:18
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    $\begingroup$ what about the $r$ from the Jacobian? :) Always remember in polar $dx \, dy$ is $r \, dr \, d\theta$. $\endgroup$
    – Math Lover
    Commented Nov 30, 2020 at 14:29
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    $\begingroup$ You can also sketch and try to see $dA$ in terms of $r, \theta$. It will be clear why $dA = r \, dr \, d\theta$. $\endgroup$
    – Math Lover
    Commented Nov 30, 2020 at 14:32
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    $\begingroup$ 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 $\theta$. $\cos 0^0 = 1, r = 3$, $\cos \pi/2 = 0, r = 0$... $\endgroup$
    – Math Lover
    Commented Nov 30, 2020 at 16:35
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    $\begingroup$ 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. $\endgroup$
    – Math Lover
    Commented Dec 1, 2020 at 10:05

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