Use an appropriate change of variables to evaluate the triple integral of (cos(x^2/4+y^2/9+z^2/25)/(sqrt(x^2/4+y^2/9+z^2/25))) dV where R is the region defined by the inequalities x^2/4+y^2/9+z^2/25<=1 and z^2/25<=x^2/4+y^2/9. u=x/2 v=y/3 w=z/5 x=2u dx=2*du y=3v dy=3*dv z=5w dz=5*dw dx dy dz=2*3*5(du dv dw) 30 triple integral of (cos(u^2+v^2+w^2)/(sqrt(u^2+v^2+w^2))) du dv dw u^2+v^2+w^2<=1 w^2<=u^2+v^2 Help me how to find the limits for the triple integral so I can find the integral.
This is what I think. I hope it isn't too late.
You have already found that the new region in terms of u, v, and w is u2 + v2 + w2 ≤ 1 and u2 + v2 ≥ w2. In words, the region is inside the sphere of radius 1 and outside the cone.
Since you have to find the integral I would strongly recommend converting into spherical coordinates from here. Doing so gives you the integral 30∫∫∫Ρcos(Ρ2)sin(Φ)dΡ dθ dΦ where 0 ≤ Ρ ≤ 1, 0 ≤ θ ≤ and (pi)/4 ≤ Φ ≤ 3(pi)/4.