# Evaluating an Integral by converting into polar coordinates.

Question. Evaluate the integral by converting into polar coordinate: $$I=\int_{0}^{\sqrt 3}\int_{0}^{\sqrt {4-y^2}}\frac{dx~dy}{4+x^2+y^2}$$

My Solution. Let $$f(x,y)=\frac{1}{4+x^2+y^2}$$. Now the region of the integration is $$S_1 \cup S_2$$ as depicted in the following figure:

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Now $$I=\iint_{S_1} f(x,y)~dxdy +\iint_{S_2} f(x,y)~dxdy=I_1+I_2.$$ Hence if I change the coordinate into polar co ordinate by $$x=r \cos \theta;~y=r\sin \theta$$ where $$0, $$I_1=\int_{r=0}^{2}\int_{0}^{\pi/3}f ~rdrd\theta$$

And $$I_2=\int_{0}^{\sqrt 3}\int_{0}^{y}f(x,y)~dxdy$$. But I cannot figure out range of $$r$$ and $$\theta$$ when $$(x,y)$$ varies in $$S_2$$.

How can I find the range of $$r$$ and $$\theta$$ in the later case? Please help. Thank you.

• In $S_2$ the angle $\theta_2$ takes all the values $\in[\pi/3,\pi/2]$. On the line segment $AB$ you have $y=\sqrt3=r\sin\theta$, so.... Commented Aug 15, 2018 at 4:37
• @JyrkiLahtonen...that means in $S_2$, $r$ varies from $0$ to $\sqrt 3/ \sin \theta$...right? Commented Aug 15, 2018 at 4:48
• Correct. ${}{}$ Commented Aug 15, 2018 at 4:50

HINT. The region $S_2$ is a right triangle. Your start angle $\theta$ clearly is the angle formed by the line $y=\sqrt{3}x$ and the $x$-axis. There is a right triangle there in your picture, I leave it to you to find that start angle. [Think, $\tan \theta$ for that right trangle.] The end angle $\theta$ is clearly vertical at $\frac{\pi}{2}$. This only leaves $r$ to take care of. This is the tricky part.

My hint for $r$ is as follows: clearly the radius depends on the angle, since different angles will force you to go different distances $r$ from the origin to 'land' on the horizontal leg formed by the region you labeled $S_2$. Draw a radius somewhere in the middle of your upper/lower $\theta$. Notice you have a right triangle inside of $S_2$: the hypotenuse being the drawn radius, and one of the legs formed by the vertical length along the $y$-axis of length $\sqrt{3}$. You can find the angle, which we shall call $\theta'$ between these two lines (since you know $\theta$). Does basic Trig give you a formula for $r$ in terms of $\theta'$? Can you replace $\theta'$ in terms of $\theta$? Then you're done!

You can always check your answer at the end: at the start angle the radius is clearly $2$ since it lies on the circle and at angle $\frac{\pi}{2}$ the radius is clearly $1$ (the leg of the right triangle formed by the region you labeled $S_2$. Does your formula for $r$ in terms of $\theta$ agree when you plug in these $\theta$?

• ....$\theta '$ should be $\pi/2- \theta$...is'n it.? Commented Aug 15, 2018 at 4:52
• Exactly right! It's the 'rest' of the angle from $\theta$ to the vertical in Quadrant $1$. Commented Aug 15, 2018 at 4:55
• And so...$r \cos \theta'=\sqrt 3$ i.e. $r=\sqrt 3/ \sin \theta$...right..? Commented Aug 15, 2018 at 4:58
• Indeed! You have $r \cos \theta' = \sqrt{3}$ so that $r=\sqrt{3}/\cos \theta'= \sqrt{3}/\cos(\pi/2-\theta)= \sqrt{3}/\sin \theta = \sqrt{3}\csc \theta$. Commented Aug 15, 2018 at 5:00
• In general, this is always the 'trick'. 'Most' of the time, $r$ will vary between two constants (never depending on $\theta$), and all is beautiful in the world. But when $r$ changes distance depending on $\theta$, the problem is a bit uglier. But once you draw a picture, somewhere there is a right triangle to help you find the relationship between the two. It's a matter of finding the appropriate triangle. Commented Aug 15, 2018 at 5:03

In $S_2$ clearly $\dfrac{\pi}{3}\leq\theta\leq\dfrac{\pi}{2}$ ans also $y\leq\sqrt{3}$ therefore $r\leq\dfrac{\sqrt{3}}{\sin\theta}$.