Surface Integral

Question

Evaluate the surface integral

∫ ∫ yzdS ,

s

where S is the part of the half-cylinder x² + y² = 9, y ≥ 0 , that lies between the planes z = 0, and x + z = 5 .

Answer 1

Oh well,

First, let's see the integration surface. It consists of 3 parts:

S1 bottom of the cylinder: z=0, x^2+y^2=9, y>=0

S2 top part of the cylinder: z=5-x, x^2+y^2=9

S3 the cylinder itself: x^2+y^2=9 and y>=0

Hence

∫ ∫ yzdS , = ∫ ∫_S1 yzdS + ∫ ∫_S1 yzdS + ∫ ∫_S3 yzdS ,

Next, we use

(*) ∫ ∫_s f(x,y,z)dS= ∫ ∫_D f(x,y,g(x,y))√(1+(dg/dx)^2+(dg/dx)^2)dA, where z=g(x,y) (in you exampe z=5-x

In the next step, you have to evaluate each of the three integrals.

For example for S2

∫ ∫_S2 yzdS= ∫ ∫_D y(5-x)√(1+(-1)^2+(0)^2)dA = √2 ∫ ∫_D y(5-x)dA , Here D={-3<=x<=3, 0<=y<=5-x)

Similar you can evaluate surface integrals for S1 and S2

OR

You can use parametrization like

x=3cost

y=3sint

z=z, 0<=t<=π

or r =3cost i +3sint j +zk, in this case ||r_txrz||=3 and

∫ ∫_S yzdS=3 ∫ ∫_D sint z dt dz=3 ∫₀ ^π sint (5-scost) dt =..

Try, if any question feel free to write me.

At least it should give you an idea how to evaluate ..