Dude i am just joining first year in IIT. I solved this question without all the triple integral nonsense. So basically we are integrating e(x^2+y^2+z^2) ^3/2 over the region x^2+y^2+z^2=1. So we make a thin sphere of thickness dr and surface area 4πr^2. So we integrate from 0 to 1 . 4πr^2 e^r^3. And get same answer
that "triple integral nonsense" is where your dV = 4 pi r^2 dr comes from, if I'd given you a cone, an ellipsoid or a quadric surface instead of a sphere you'd be struggling
Dude i am just joining first year in IIT.
I solved this question without all the triple integral nonsense.
So basically we are integrating e(x^2+y^2+z^2) ^3/2 over the region x^2+y^2+z^2=1.
So we make a thin sphere of thickness dr and surface area 4πr^2. So we integrate from 0 to 1 . 4πr^2 e^r^3. And get same answer
that "triple integral nonsense" is where your dV = 4 pi r^2 dr comes from, if I'd given you a cone, an ellipsoid or a quadric surface instead of a sphere you'd be struggling
Cylindrical coordinates.