Answer:
9828 coulombs
Explanation:
z = [tex]\sqrt{3x^2 + 3y^2}[/tex]
Relation between the cartesian and polar coordinates
x = r cos ∅
y = r sin ∅
z = [tex]\sqrt{3x^2 + 3y^2}[/tex] = [tex]\sqrt{3r}[/tex]
next we calculate the surface charge density of the cone
б ( r, ∅ ) = xyz^2 = 3r^4 sin∅ cos∅
Charge given as
Q = ∫ б(r,∅ )ds
= ∫ б(r,∅ ) [tex]\sqrt{1 +( \frac{dz}{dx})^2 + (\frac{dz}{dx} )^2 } dA[/tex]
after integration
Q = ∫ б(r,∅ ) 2dA
Attached below is the remaining part of the solution
