Calculate the double integral of F.ndS?

Calculate the double integral of F.ndS where F = (ax^3)i + (by^3)j +(cz^3)k


(where a, b and c are constants) over the surface of a sphere of radius r centred at the origin


and over the surface of a cylinder of radius r and height 2h centred at the origin with its axis in


the z-direction

Calculate the double integral of F.ndS?
Ok, F = (ax³)i + (by³)j +(cz³)k





The general algorithm to calculate ∫∫F.ndS is to determine:


i) the (possibly piecewise) equation in x,y,z that parameterizes the entire surface


ii) n, the normal vector at each point on that surface. Usually got from the del(gradient) operator on the surface equation from i).


iii) dS, the infinitesimal area of a patch at point (x,y,z) on the surface.


In rectangular/Cartesian coordinates this is:


dS = dx dy / |n . k| (for explanation, see text link below)


or dS = dx dz / |n . j| if you are parameterizing for x,z


For the sphere this may be easier to formulate in spherical coordinates.


iv) finally compute the general value


F.ndS = F.n dx dy / |n . k|


v) and compute the double integral (piecewise if necessary) over the entire surface





SPHERE of radius r:


i) Eqn is: x² + y² + z² = r, i.e F(x,y,z) = x² + y² + z² - r = 0


for -r%26lt;=x%26lt;=r, -r%26lt;=y%26lt;=r


Thus there are two values of z, each has to be considered:


z² = r - x² + y²


z = +/-√(r - x² + y²)





Note: del F(x,y,z) = (2x, 2y, 2z)


ii) n = del F(x,y,z) / | del F(x,y,z) |


= (x,y,z) / r = 1/r (x,y,z)


iii)


dS = dx dy / |n . k|


= dx dy / |(x,y,z) / r . k|


= dx dy / |z/r|


= r dx dy / |z|





iv)


F.n = (ax³,by³,cz³) . 1/r (x,y,z)


= 1/r (ax^4 + by^4 + cz^4)


Then:


F.n dS = 1/r (ax^4 + by^4 + cz^4) * r dx dy / |z|


= (ax^4 + by^4 + cz^4) dx dy / |z|





We can eliminate z for x,y with:


z² = r - x² + y²


z = √(r - x² + y²)


Thus:


F.n dS = (ax^4 + by^4 + c(r - x² + y²)²) / √(r - x² + y²) dx dy





v)


∫∫ F.n dS from x=-r..+r, y=-r..+r


= ∫∫ (ax^4 + by^4 + c(r - x² + y²)²) / √(r - x² + y²) dx dy


= ...





CYLINDER of radius r, height 2h centred at the origin with its axis in the z-direction:


i) Piecewise parameterization is the curved surface, and top and bottom circles.





Curved surface:


F(x,y) = x² + y² = r, z goes from -h to +h


Thus: -h%26lt;=z%26lt;=+h, y = +/- √(r - x²)


F = (x,√(r - x²),z)





Top and bottom circles:


x² + y² %26lt;= r, z= +/-h


So: -r%26lt;=x%26lt;=r, -h%26lt;=z%26lt;=+h


-√(r - x²)%26lt;=y%26lt;=+√(r - x²)





ii) Unit Normal n = del F(x,y,z) / | del F(x,y,z) |





Curved surface: F = (x,√(r - x²),0)


del F = (1, x/√(r - x²),0)


n = (1, x/√(r - x²),0) / | del F(x,y,z) |


n = 1/r (√(r - x²), x,0)





Top circle: n = (0,0,1)


Bottom circle: n =(0,0,-1)





iii)


Curved surface: dS = dx dz / |n . j|


dS = 1/r x dx dz





Top and bottom circles:dS = dx dy / |n . k|


=%26gt; dS = dx dy





iv)


Curved surface:


F.n = (ax³,by³,cz³) . 1/r (√(r - x²), x,0)


= (ax³,b(r - x²)^(3/2),cz³) . 1/r (√(r - x²), x,0)


= 1/r (ax³√(r - x²) + bx(r - x²)^(3/2))





F.n dS = 1/r (ax³√(r - x²) + bx(r - x²)^(3/2)) 1/r x dx dz


F.n dS = 1/r² (ax^4√(r - x²) + bx²(r - x²)^(3/2)) dx dz





Top circle:


F.n dS = (ax³,by³,cz³).(0,0,1) dx dy


F.n dS = cz³ dx dy





etc.
Reply:here's the first one





n=2xi+2yj+2ak/(2r)=x/ri+y/rj+z/rk





∫∫F∙ndxdy=∫∫(ax^3i+by^3+cz^3)∙(x/ri+y/... dx=





∫∫(ax^4i+by^4+cz^4)/r dx dy


replace z with sqrt(r^2-x^2-y^2)





∫∫(ax^4i+by^4+c((r^2-x^2-y^2)^2)dx dy





the limits of integration for


y are -sqrt(r^2-x^2) to sqrt(r^2-x^2) the limits for x are -r to r.





The calculations from here are straightforward.