Question

a) find center of mass of a solid of constant density bounded below by the paraboloid z=x^2+y^2 and above by the plane z=4.(b) Find the plane z=c that divides the solid into two parts of equal volume. This plane does not pass through the center of mass.The answers in the back of the book are:(a) x-bar=y-bar=z-bar=(8/3)(b) c=2(sqrt(2))

Answer 1

Answer: x-bar = y-bar = 0 whereas z-bar = 8/3

               M= (c^2)/8 which is intern equal to 2$\sqrt{2}$

Step-by-step explanation:

           Find the area, by setting the limits as

               = $4\cdot \int _0^{\frac{\pi }{2}}\int _0^2\int _{r^2}^4\:rdzdrd\theta$

                $=4\cdot \int _0^{\frac{\pi }{2}}\int _0^2r\cdot \:4-r^3drd\theta$

                $=4\cdot \int _0^{\frac{\pi }{2}}4d\theta$

                 $=8\pi$

Therefore;

Mxy=  $\int _0^{2\pi }\int _0^2\int _{r^2}^4\:zrdzdrd\theta$

       z-bar = 8/3

M= 8$\pi$ dividing it into two volume gives us = 4$\pi$

means  $4\pi =\int _0^{2\pi }\int _0^{\sqrt{c}}\int _r^c\:rdzdrd\theta$

             $4\pi =\left(\pi c^2-2\pi \frac{c^{\frac{3}{2}}}{3}\right)$

              c=2$\sqrt{2}$