How do you find the volume of the solid obtained by revolving the curve given by #x=3cos^3(t)#, #y=5sin^3(t)# about the #x#-axis?

Redirected from "How do I change #int_0^1int_0^sqrt(1-x^2)int_sqrt(x^2+y^2)^sqrt(2-x^2-y^2)xydzdydx# to cylindrical or spherical coordinates?"
1 Answer
Sep 20, 2014

#y^2=25sin^6t=25(1-cos^2t)^3=25[1-(x/3)^{2/3}]^3#

By Disk Method,

#V=pi int_{-3}^3y^2 dx=25piint_{-3}^3[1-(x/3)^{2/3}]^3dx#