[最新] z=sqrt(x^2 y^2) in polar coordinates 884244

Problem Easy Difficulty Use polar coordinates to find the volume of the given solid Below the cone $ z = \sqrt{x^2 y^2} $ and above the ring $ 1 \le x^2 y^2 \le 4 $

Z=sqrt(x^2 y^2) in polar coordinates-Problem 26 Medium Difficulty Use spherical coordinates Evaluate $ \iiint_E \sqrt{x^2 y^2 z^2}\ dV $, where $ E $ lies above the cone $ z = \sqrt{x^2 y^2} $ and between the spheres $ x^2 y^2 z^2 = 1 $ and $ x^2 y^2 z^2 = 4 $Last, in rectangular coordinates, elliptic cones are quadric surfaces and can be represented by equations of the form z 2 = x 2 a 2 y 2 b 2 z 2 = x 2 a 2 y 2 b 2 In this case, we could choose any of the three However, the equation for the surface is more complicated in rectangular coordinates than in the other two systems, so we might