Question

Compute the electric charge on the surface which is the portion of the cone z = 3x2 + 3y2 in the first octant that lies between the planes z = 2 and z = 6. The charge density on the surface is given by σ(x, y, z) = xyz2 coulombs per square meter.

Answer 1

Answer:

9828 coulombs

Explanation:

z = $\sqrt{3x^2 + 3y^2}$

Relation between the cartesian and polar coordinates

x = r cos ∅

y = r sin ∅

z = $\sqrt{3x^2 + 3y^2}$ = $\sqrt{3r}$

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,∅ ) $\sqrt{1 +( \frac{dz}{dx})^2 + (\frac{dz}{dx} )^2 } dA$

after integration

Q  = ∫ б(r,∅ ) 2dA

Attached below is the remaining part of the solution