Integration with Polar Coordinates

Question

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.

Answer 1

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.