How do I figure out the integral bonds for r when integrating over the area M under a parabola y=2-x^2?
Do I need to solve a quadratic equation? I think there must be an easier solution.
Clarification
I want to calculate the following integral using polar coordinates (in this order).
$\iint_Mf(r,\varphi)drd\varphi$
$\endgroup$ 41 Answer
$\begingroup$So I assume that you have an integrand which is awkward in cartesian coordinates but OK in polar coordinates so that you want to perform the integration in polar coordinates. At the boundary, $r\sin\phi=2-r^2\cos^2\phi$ so by the quadratic formula$$\begin{align}r&=\frac{-\sin\phi\pm\sqrt{\sin^2\phi+8\cos^2\phi}}{2\cos^2\phi}\\ &=\frac{\sin^2\phi-\sin^2\phi-8\cos^2\phi}{2\left(-\sin\phi\mp\sqrt{\sin^2\phi+8\cos^2\phi}\right)\cos^2\phi}=\frac4{\sin\phi+\sqrt{\sin^2\phi+8\cos^2\phi}}\end{align}$$Where we have taken the upper signs because we want $r>0$. Thus$$\int_{-\sqrt2}^{\sqrt2}\int_0^{2-x^2}f(x,y)\,dy\,dx=\int_0^{\pi}\int_0^{\frac4{\sin\phi+\sqrt{\sin^2\phi+8\cos^2\phi}}}f(r\cos\phi,r\sin\phi)r\,dr\,d\phi$$
$\endgroup$ 3
