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$D=\{(x,y)\mid x^2+y^2 \le x\}$ , by changing to polar coordinate, find this integral: $$\iint_{D}\sqrt x\,dx\,dy$$

So I got this by changing to polar coordinate

$$\int_{-\pi/2}^{\pi/2}\int_{0}^{1/2}\sqrt{r\cos \theta}\ rdr\,d\theta$$
First is my limit for integral true?

And I don't know how to integrate $\sqrt {r\cos\theta} $ here.

Thank you!

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    $\begingroup$ Changing to polars means $dxdy = r dr d\theta$. You're missing an $r$. You've also made multiple edits now which has changed the nature of the question. Make sure the question is correct as you have stated it. $\endgroup$
    – Matthew Cassell
    Aug 25, 2017 at 5:35
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    $\begingroup$ You might want to think a bit more carefully about the domain. The one you have written in the double integral is a semicircle centered on the origin. However, when we write $D$ like so $$D=\left\{(x, y)\ |\ \left(x-\frac 12\right)^2+y^2\le\frac 14\right\}$$ We see it should be a circle centered at $(\frac 12, 0)$ $\endgroup$
    – Kajelad
    Aug 25, 2017 at 5:36
  • $\begingroup$ @kajelad but the circle is in first and fourth quadrant (?) or it should be $0-> 2 \pi$? $\endgroup$
    – fiksx
    Aug 25, 2017 at 5:41
  • $\begingroup$ If you let $x=\frac{1}{2}+r\cos\theta $ and $y=r\sin\theta$ you don't have to worry about quadrants and you get $$\int_{0}^{2\pi} \int_{0}^{1/2}r\sqrt{\frac{1}{2}+r\cos\theta}\,dr\,d\theta$$ $\endgroup$
    – Thomas Andrews
    Aug 25, 2017 at 5:44
  • $\begingroup$ If the circle isn't centered at the origin, writing the bounds of integration won't be so simple. You can, however, apply a change of coordinates so that your new origin is at $(\frac 12, 0)$ , then the bounds will be very neat (as Thomas Andrews demonstrates above). Alternately, you can just apply the change of coordinates $x=r\cos(\theta),\ y=r\sin(\theta)$ directly to the inequality. $\endgroup$
    – Kajelad
    Aug 25, 2017 at 5:44

1 Answer 1

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Converting to polar we see $D= \{(r, \theta) | r^2 - r\cos\theta \le 0\} =\{(r,\theta) | r\le \cos\theta\}$.

Not forgetting to multiply by the Jacobian (which I suspect you did originally) we have that

$$\int \int_D \sqrt{x}dxdy = \int_{-\pi/2}^{\pi/2}\int_0^{\cos\theta}r\sqrt{r}\sqrt{\cos\theta}drd\theta$$ $$ = \int_{-\pi/2}^{\pi/2}\frac{2}{5}\cos^3\theta d\theta = \frac25 \int_{-\pi/2}^{\pi/2}\cos\theta (1 -\sin^2\theta)d\theta$$ $$ = \frac25 \left(\sin\theta - \frac13 \sin^3\theta \biggm|_{-\pi/2}^{\pi/2}\right)= \frac25 \cdot \frac43 = \frac{8}{15}.$$

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  • $\begingroup$ thankyou! but can you explain to me why $ \int_{-\pi/2}^{\pi/2}$? $\endgroup$
    – fiksx
    Aug 25, 2017 at 6:02
  • $\begingroup$ Can the radius of a vector be less than the cosine of the angle it makes with the positive $x$-axis is outside those bounds? More precisely: What is $\cos\theta$ for $\theta \in [-\pi , \frac\pi2] \cup [\frac\pi2, \pi]$? Can it ever be true that $r<\cos\theta$? $\endgroup$
    – David Bowman
    Aug 25, 2017 at 6:18

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