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I thought $dA=\sin\theta d\theta d\phi$? What did I missed?


@OtB_BlueBerry It depends on how I parametrize by integral. For example, if I were to re-parametrize my integral by setting $\theta = \pi/2 - x$, then I can rewrite my integral equivalently as follows: $\int_0^{2\pi} \int_0^{\pi} g(x,\phi) sin(x)~dx~d\phi$ where $g(x,\phi) = f(\pi/2 - x,\phi)$. And voila!