1. The problem statement, all variables and given/known data Set up triple integrals for the integral of f(x,y,z)=6+4y over the region in the first octant that is bounded by the cone z=(x^2+y^2), the cylinder x^2+y^2=1 and the coordinate planes in rectangular, cylindrical, and spherical coordinates. 2. Relevant equations ∫∫∫dzrdrdθ ∫∫∫[itex]\rho[/itex]^2*sin[itex]\phi[/itex]d[itex]\rho[/itex]d[itex]\phi[/itex]d[itex]\theta[/itex] ∫∫∫dxdyzdz 3. The attempt at a solution I think this problem is mainly giving me a hard time due to that it is in the first octant. I'm still having a hard time visualizing things in three dimensions. First thing I did was sketch the region. I have the cylinder with a radius of 1 centered on the z-axis. My cone, starts at the origin, and extends to a maximum radius of 1 where it touches the cylinder. (Now this is where I get a little confused...) The axes all bound the region as well, so the z axis cuts the cone and the cylinder in half, and then the half remaining is once again cut in half by the x and y axes? Ok. So given that's at least remotely correct, I began by setting up the integral in cylindrical coordinates. For my z limits of integration, I said from z=(x^2+y^2)^1/2, which is z=r to z=1. For my r limits, from 0 to 1. And I'm confused about [itex]\theta[/itex], since we're dealing with the "first octant," so I'm thinking it's from 0 to pi/2. My function itself, 6+4y, is not in the appropriate coordinate system. Does the function in the integral become 6+4(rsin[itex]\theta[/itex])? For my cartesian coordinates, I said that my z limits of integration should be the same, I think. Now when I look at the circle on the xy-plane, since we're looking in only one octant, I'm essentially looking at a quarter of the circle, right? So my x-limits are from x=0 to x=(1-y^2)^1/2. And my y limits are simply 0 to 1. Lastly, spherical coordinates. For my limits for [itex]\rho[/itex], I have from 0 to 1. [itex]\phi[/itex] is from 0 to pi/4, and [itex]\theta[/itex] is from 0 to pi/2. Now as far as the function inside the integral, I already have 6+4y, which converts to 6+4([itex]\rho[/itex]sin[itex]\phi[/itex]sin[itex]\vartheta[/itex]) and then multiplied by [itex]\rho[/itex]^2*sin[itex]\phi[/itex]. Whew. Alright. Hopefully I'm somewhat on the right track, at least. Hope my formatting is easy enough to follow. Thanks so much!