Find the mass of the conical surface z=sqrt(x^2+y^2) for z≤ 2 with density p(x,y)=x^2+y^2 .



any help will be greatly appreciated

I suppose is a solid cone , finally i will take the surface only :-)

Since x^2+y^2=z^2 , at height z you have a cross section of a circle with radius R^2=z^2
ie , R=z .- You have a disk parallel to XY plane at z, with radius R=z .- The mass of an element of this disk is dm= p(x,y) dV

take cylindricals coordinates on this disk , x=rcosT , y=rsinT , z=dz
p(x,y) =p(r,T) = r^2

dV= dA dz = (rdT dr) dz
dm=( r^3 dr dT) dz

M= INT INT INT r^3 dr dz dT 0<r<z 0<T<2pi 0<z<2
M= 2pi INT (z^4/4) dz
M= (pi/2) INT z^4 dz

M= (pi/10) z^5 = 32pi/10

If it was the mass of the surface only , i think is better to use a surface integral .-
dm= p(x,y) dS dS Element of surface

dS= dydx / I n dot k I it is the surface dS projected on XY plane , so the Region will be the region at z=2 , ie x^2+y^2=4 ( A circle with radius R=2 )

Take polar coordinates , x(r,T) =rcosT , y(r,T) =rsinT z= r ( Cone) , thus

A position vector of the cone is R(r,T) = rcosT i + rsinT j + r k
A normal vector is N= (dR/dr) x (dR/dT)
dR/dr = cosT i +sinT j + k
dR/dT = -rsinT i +rcosT j + 0k

N= -rcosT i -rsinT j + r k
INI = r sqrt ( 2 )

n = N/INI and
(n dot k ) = 1/sqrt2
dydx = dA= rdT dr
p(x,y)=x^2+y^2 .= r^2

dm= sqrt 2 (r^2 (rdr dT ) )

dm= sqrt2 r^3 dr dT 0<r<2 , 0<T<2pi

M= 2pi sqrt2 (r^4/4)
M= 8pi sqrt2 Wonderful !!