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I want to integrate this integral with polar coordinates:

$\int \sin x \ dA$ on the region bounded by $ y=x, y=10-x^2, x=0$.

So far I've got that $$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \int_{l}^{10} f(r\cos\theta, r\sin\theta) \ r\,dr\,d\theta$$

Where $l =\sqrt{2}\frac{(\sqrt{41}-1)}{2}$.

I'm most curious about how I should represent $\sin x$, though I could directly use $\sin(r\cos\theta)$. I feel like there's a better way though.

Thx for helping.

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I would suggest against polar coordinates for two reasons:

  • The transformation is indeed $\sin(x)=\sin(r\cos\theta)$, which will be a beast to integrate.
  • Polar coordinates are to your advantage with circular bounds. You don't have that here. You would if it was $y=\sqrt{10-x^2}$, but that is not the case.

I would stick to cartesian coordinates for this integral. It still won't be nice, but it will be better. You are going to want to look up the Fresnel integrals.

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  • $\begingroup$ Thx. Your second point I considered before as a typo in the question, but I haven't had the liberty of asking the instructor. Is my polar conversion correct then? And then I should evaluate $\int_{0}^{4} \int_{y}^{\sqrt{10-y}} \sin x \ dxdy$ instead (i.e., is that the correct rectangular form)? $\endgroup$ – Joshua Bunce Jun 26 '15 at 5:36
  • $\begingroup$ The $4$ needs to be $10$ $\endgroup$ – Joshua Bunce Jun 26 '15 at 5:43
  • $\begingroup$ Sorry, that is not correct. If you integrate in $x$ first, you will have to break it up into a sum of two integrals because the bounds on $x$ have different forms at different $y$. $\endgroup$ – Alex S Jun 26 '15 at 5:43
  • $\begingroup$ Not sure I'm understanding. Is not $x$ bounded between the line $y=x$ and the parabola $y=10-x^2$? graph of region $\endgroup$ – Joshua Bunce Jun 26 '15 at 5:48

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