
#1
Nov1508, 07:36 PM

P: 9

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=4x^{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=4r^{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 4r^{2} to 3r^{2}, so that gave me the equation: [tex]\int^{2\pi}_0\int^1_{1}\int^{3r^2}_{4r^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 zlimits around, so now it is: [tex]\int^{2\pi}_0\int^1_0\int^{4r^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. 



#2
Nov1608, 04:07 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,894

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]. 


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