1. The problem statement, all variables and given/known data Use polar coordinates to find the volume bounded by the paraboloids z=3x^{2}+3y^{2} and z=4-x^{2}-y^{2} 2. Relevant equations 3. The attempt at a solution Somehow, through random guessing, I managed to get the right answer, it's just that I don't understand how I got it. Also, because the z is involved, I actually used cylindrical coordinates, but would that still be considered the same thing as polar coordinates? So anyway, I changed the two paraboloid equations to z=3r^{2} and z=4-r^{2}. Then setting these two equations equal to each other (since they are both equal to z), I solved for r and got the limits of -1,1. For the limits of theta, I just happened to take it from 0 to 2pi. Lastly for the z limits, I just tried from 4-r^{2} to 3r^{2}, so that gave me the equation: [tex]\int^{2\pi}_0\int^1_{-1}\int^{3r^2}_{4-r^2}dzrdrd\theta[/tex] However, solving this equation didn't give me the right answer, so I changed the limits of r to 0 to 1, and switched the z-limits around, so now it is: [tex]\int^{2\pi}_0\int^1_0\int^{4-r^2}_{3r^2}dzrdrd\theta[/tex] And solving for this, gave me the right answer of 2pi. The problem is that I don't understand the real logic behind what I did. So in summary, what I didn't understand was how to establish the limits for theta, r, and z.
Those two paraboloids intersect when z= 3x^{2}+ 3y^{2}= 4- x^{2}- y^{2} or 4x^{2}+ 4y^{2}= 4 which reduces to x^{2}+ y^{2}= 1. That projects onto the xy-plane as the unit circle. It shouldn't be hard to see that the volume you want is above that circle. To cover that circle, [itex]\theta[/itex] goes from 0 to [itex]2\pi[/itex] and r goes from 0 to 1. Those are your limits of integration. For each (x, y), the upper boundary is z= 4- x^{2}- y^{2} and the lower boundary is z= 3x^{2}+ 3y^{2}. The volume of a "thin" rectangular solid used to construct the Riemann sums for this volume would be [(4- x^{2}- y^{2})- (3x^{2}+ 3y^{2})]dxdy= (4- 4x^{2}- 4y^{2})dxdy= 4(1- x^{2}- y^{2})dxdy and that, in polar coordinates, is 4(1- r^{2})rdrd[itex]\theta[/itex]. Yes, "cylindrical coordinates" is just polar coordinates in the xy-plane. No, you didn't. Or you shouldn't. r cannot be negative. r must be between 0 and 1. Did you think at all about the geometry of the situation? This has circular symmetry. To cover the entire volume you have to cover the entire circle: 0 to [itex]2\pi[/itex]. Surely you recognized that z= 4- r^{2} is ABOVE z= 3r^{2} for r<= 1?