#Integral Triple
21 messages · Page 1 of 1 (latest)
Is the first pic the hyperboloid function?
the first pic is the formula of the lateral surface
How I always think about these integrals is by breaking it down into steps. Starting with the simplest
You want to integrate for mass, so it's literally just $M=\int dm$. Then we need to know the bounds of $m$. The bounds are spatial ($dV$). So how do we related mass and volume? That's where density comes in. We know the density is constant in the hyperboloid cylinder, and 0 outside. Let the density constant be $\rho$. Then $dm=\rho dV$.
SWR
So the integral becomes $M=\iiint \rho dV$, bounded by the hyperboloid.
SWR
You are asked to integrate in cylindrical coordinates, which makes sense given the nice symmetry
Do you remember $dV$ in cylindrical coordinates?
SWR