Question

Find the mass and center of mass of the solid E with the given density rho. E is the cube 0 ≤ x ≤ a, 0 ≤ y ≤ a, 0 ≤ z ≤ a; rho(x, y, z) = 9x2 + 9y2 + 9z2.

Answer 1

Answer:

mass = 9a^5

center of mass = $\frac{7a}{12}, \frac{7a}{12}, \frac{7a}{12}$

Step-by-step explanation:

Finding the mass of the solid E

given density function : p ( x,y,z ) = $9x^2 + 9y^2 + 9z^2$

Mass =   $\int\limits^a_0 \int\limits^a_0 \int\limits^a_0 {9(x^2+y^2+z^2)} \, dx dydz$   $= \int\limits^a_0 \int\limits^a_0 {9(\frac{a^3}{3}+ay^2+az^2 )} \, dydz$

         $= \int\limits^a_0 {9(\frac{a^4}{3}+\frac{a^4}{3} +a^2z^2 )} \, dz$  $= \int\limits^a_0 {9(\frac{2a^4}{3}+a^2z^2 )} \, dz$  $= 9 ( \frac{2a^5}{3} + \frac{a^5}{3} )$  

( taking limits as a and 0 )

hence Mass = 9 $(a^5)$

finding the center of mass

attached below is solution