Question

Find the volume (triple integral)

Answer 1

Find the volume of the solid region that lies above the surface \[z = 1/2(x^2+y^2)\] and below the surface \[z = \sqrt{x^2+y^2}\]

@ganeshie8

Answer 2

So it's a triple integral right, so z = 1/2(r^2), and z = r
Setting z = z we get r = 2 so radius is [0, 2]
theta [0, 2pi]
z [1/2r^2, r] \[\large \int\limits_{0}^{2 \pi}\int\limits_{0}^{2} \int\limits_{1/2r^2}^{r} rdzdrd \theta\] this good?