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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