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I'm trying to convert the bounds of integration to polar coordinates but I'm stumped on one of the bounds.

$$\int_{x=0}^{6}\int_{y=\frac{1}{\sqrt{3}}x}^{\sqrt{8x-x^2}}\sqrt{x^2+y^2}\,dy\,dx$$

The only thing that left me stumped was converting $y=\frac{1}{\sqrt{3}}x$ to polar.

Right now I have $\int_{\theta=0}^{\frac{\pi}{6}}\int_{?}^{8\cos{\theta}}r^2\,dr\,d\theta$

Where do I go from here? It's nowhere in my notes and I'm having a tough time finding anything online about it. Thank you!

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3 Answers 3

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For $r,$ your lower bound is simply zero. Please sketch and see.

enter image description here

You have a circle with radius $4$ at center $(4,0)$. You are integrating over the region above the line $y = \frac {x} {\sqrt3}$ and below the circle. In polar form the radius will go from zero to $8 \cos \theta$. The bound on $\theta$ ensures you are integrating over the circular segment.

EDIT: I just noticed bound of your $\theta$. That has to be from $\pi/6 \,$ to $\pi/2$.

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  • $\begingroup$ Thank you! Just clicked in my head all the sudden and makes sense. $\endgroup$
    – Isaackoz
    Nov 5, 2020 at 21:46
  • $\begingroup$ say if the bounds were from $-4 \le y \le 4$ and $0 \le x \le \sqrt{16 - y^2}$ ?? $\endgroup$ Dec 11, 2020 at 5:27
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Let's do a quick plot: enter image description here

Now it should be obvious that $\theta$ goes from $\frac\pi6$ to $\frac\pi2$ and $r$ goes from $0$ to $8\cos\theta$

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  • $\begingroup$ say if the bounds were from $-4 \le y \le 4$ and $0 \le x \le \sqrt{16 - y^2}$ ?? $\endgroup$ Dec 11, 2020 at 5:28
  • $\begingroup$ @user1618033988749895 Make a picture. This is half a disk, of radius $4$, right of the $y$-axis ($x=0$). Can you figure out the limits now? $\endgroup$
    – Andrei
    Dec 11, 2020 at 9:18
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The domain of integration is delimited by the line with equation $y=\frac1{\sqrt 3}$ and the part of the circle with centre $(4,0)$ and radius $4$ since $$y=\sqrt{8x-x^2}\iff y^2+x^2-8x=0,\; y\ge 0\iff (x-4)^2+y^2-16=0,\;y\ge 0.$$

On the other hand, by the law of sines, we have the relation $$\frac r{\sin2\theta}=\frac4{\sin\theta}\quad\text{whence the polar equation of the circle}\quad r=\frac{4\sin2\theta}{\sin\theta}=8\cos\theta,$$ and the integral becomes $$\int_{\tfrac\pi6}^\tfrac\pi2\int_0^{8\cos\theta}\mkern-12mu r^2\,\mathrm d r\,\mathrm d\theta.$$

enter image description here

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  • $\begingroup$ say if the bounds were from $-4 \le y \le 4$ and $0 \le x \le \sqrt{16 - y^2}$ ?? $\endgroup$ Dec 11, 2020 at 5:28
  • $\begingroup$ Well, it's clearly the half-disk of radius $4$ centred at the origin, on the right of the $y$-axis, so $r$goes from $0$ to $4$ and $\theta$ from $-\dfrac\pi2$ to $\dfrac\pi 2$. $\endgroup$
    – Bernard
    Dec 11, 2020 at 9:54

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