How do I solve this integral? This is the center of gravity of a cone.

#Z_S= int_0^hint_0^(2pi)int_0^(R(1-z/h)) mu_0 z rho^2 drho dvarphidz#

1 Answer
Mar 25, 2018

#=>Z_S = (pi mu_0 R^3h^2)/30#

Explanation:

We start with:

#Z_S= int_0^hint_0^(2pi)int_0^(R(1-z/h)) mu_0 z rho^2 drho dvarphidz#

We can pull out the #mu_0#:

#= mu_0 int_0^hint_0^(2pi)int_0^(R(1-z/h)) z rho^2 drho dvarphidz#

We integrate w.r.t. #rho#:

#= mu_0 int_0^hint_0^(2pi) z [rho^3/3]_{0}^{R(1-z/h)} dvarphidz#

We assess the limits for #rho#:

#= mu_0/3 int_0^hint_0^(2pi) z (R(1-z/h))^3 dvarphidz#

#= mu_0/3 int_0^hint_0^(2pi) z R^3(1-z/h)^3 dvarphidz#

#= (mu_0 R^3)/3 int_0^hint_0^(2pi) z(-z^3/h^3 + (3 z^2)/h^2 - (3 z)/h + 1) dvarphidz#

#= (mu_0 R^3)/3 int_0^hint_0^(2pi) (-z^4/h^3 + (3 z^3)/h^2 - (3 z^2)/h + z) dvarphidz#

Now we integrate with w.r.t. #varphi#:

#= (mu_0 R^3)/3 int_0^h 2pi(-z^4/h^3 + (3 z^3)/h^2 - (3 z^2)/h + z) dz#

Lastly we integrate w.r.t. #z#:

#= (2pi mu_0 R^3)/3 int_0^h (-z^4/h^3 + (3 z^3)/h^2 - (3 z^2)/h + z) dz#

We assess the limits for #z#:

#= (2pi mu_0 R^3)/3 [(-z^5/(5h^3) + (3 z^4)/(4h^2) - (z^3)/h + z^2/2)]_{0}^{h}#

#= (2pi mu_0 R^3)/3 (-h^5/(5h^3) + (3 h^4)/(4h^2) - (h^3)/h + h^2/2)#

#= (2pi mu_0 R^3)/3 (-h^2/(5) + (3 h^2)/(4) - h^2 + h^2/2)#

#= (2pi mu_0 R^3)/3 (-(4h^2)/(20) + (15 h^2)/(20) - (20h^2)/20 + (10h^2)/20)#

#= (2pi mu_0 R^3)/3 (h^2/20)#

#= (2pi mu_0 R^3h^2)/60#

#= (pi mu_0 R^3h^2)/30#

Hence:

#=>color(green)(Z_S = (pi mu_0 R^3h^2)/30)#