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Section 17.6 : Divergence Theorem

3. Use the Divergence Theorem to evaluate SFdS where F=2xzi+(14xy2)j+(2zz2)k and S is the surface of the solid bounded by z=62x22y2 and the plane z=0 . Note that both of the surfaces of this solid included in S.

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Start Solution

Let’s start off with a quick sketch of the surface we are working with in this problem.

We included a sketch with traditional axes and a sketch with a set of “box” axes to help visualize the surface. The bottom “cap” of the elliptic paraboloid is also included in the surface but isn’t shown.

Note as well here that because we are including both of the surfaces shown above that the surface does enclose (or is the boundary curve if you want to use that terminology) the region.

Show Step 2

We are going to use Divergence Theorem in the following direction.

SFdS=EdivFdV

where E is just the solid shown in the sketches from Step 1.

The region D for that we’ll need in converting the triple integral into iterated integrals is just the intersection of the two surfaces from the problem statement. This is,

0=62x22y2x2+y2=3

So, D is a disk and so we’ll eventually be doing cylindrical coordinates for this integral. Here are the cylindrical limits for the region E.

0θ2π0r30z62x22y2=62r2

Don’t forget to convert the z limits into cylindrical coordinates as well!

We’ll also need the divergence of the vector field so here is that.

divF=x(2xz)+y(14xy2)+z(2zz2)=28xy Show Step 3

Now let’s apply the Divergence Theorem to the integral and get it converted to cylindrical coordinates while we’re at it.

SFdS=EdivFdV=E28xydV=2π03062r20(28r2cosθsinθ)rdzdrdθ=2π03062r202r8r3cosθsinθdzdrdθ

Don’t forget to pick up the r when converting the dV to cylindrical coordinates.

Show Step 4

All we need to do then in evaluate the integral.

SFdS=2π03062r202r8r3cosθsinθdzdrdθ=2π030(2r8r3cosθsinθ)z|62r20drdθ=2π030(2r8r3cosθsinθ)(62r2)drdθ=2π03012r4r3(48r316r5)cosθsinθdrdθ=2π0(6r2r4(12r483r6)cosθsinθ)|30dθ=2π0936cosθsinθdθ=2π0918sin(2θ)dθ=(9θ+9cos(2θ))|2π0=18π