- #1
haroldholt
- 21
- 0
The question I was given asks to verify the divergence theorem by showing that both sides of the theorem show the same result. With the divergence theorem obviously being \iint_S\mathbf{F}\cdot\mathbf{n}\,dS = \iiint_V \nabla\cdot\mathbf{F}\,dV.
The vector field is \mathbf{F}=\mathbf{i}+\mathbf{j}+\mathbf{k} and the surface through which it is flowing is the cone z=\sqrt{x^2+y^2} below the plane z = 1.
Firstly, as far as I can tell, \iiint_V \nabla\cdot\mathbf{F}\,dV = 0 since \nabla\cdot\mathbf{F} = 0 which covers the right hand side of the divergence theorem.
So then I divided the surface into two, that is the top of the cone and the rest of the cone.
For the top of the cone I got the surface integral as \frac{\pi}{2} since \mathbf{F}\cdot\mathbf{n} = (\mathbf{i}+\mathbf{j}+\mathbf{k})\cdot(\mathbf{k}) = 1 so then you're just left with the surface integral of 1 which by definition is the area of the surface. And since the radius is \sqrt{\frac{1}{2}}, the area is \frac{\pi}{2}.
Now for the other surface I found the unit normal to be \frac{\cos{t}+\sin{t}-1}{\sqrt{2}} after parametrizing the surface as \mathbf{r} = z\cos{t}\mathbf{i} + z\sin{t}\mathbf{j} + z\mathbf{k} thus finding \iint_S\mathbf{F}\cdot\mathbf{n}\,dS = -\sqrt{2}\pi.
Obviously I wanted the result to be -\frac{\pi}{2} so that the total surface integral would be equal to zero which would verify the divergence theorem for this problem. But it's not and I can't figure out where I went wrong. Since -\sqrt{2}\pi is fairly close to -\frac{\pi}{2} I figure I'm on the right track at least with my working.
Hope this makes sense and thank you in advance for any assistance.
The vector field is \mathbf{F}=\mathbf{i}+\mathbf{j}+\mathbf{k} and the surface through which it is flowing is the cone z=\sqrt{x^2+y^2} below the plane z = 1.
Firstly, as far as I can tell, \iiint_V \nabla\cdot\mathbf{F}\,dV = 0 since \nabla\cdot\mathbf{F} = 0 which covers the right hand side of the divergence theorem.
So then I divided the surface into two, that is the top of the cone and the rest of the cone.
For the top of the cone I got the surface integral as \frac{\pi}{2} since \mathbf{F}\cdot\mathbf{n} = (\mathbf{i}+\mathbf{j}+\mathbf{k})\cdot(\mathbf{k}) = 1 so then you're just left with the surface integral of 1 which by definition is the area of the surface. And since the radius is \sqrt{\frac{1}{2}}, the area is \frac{\pi}{2}.
Now for the other surface I found the unit normal to be \frac{\cos{t}+\sin{t}-1}{\sqrt{2}} after parametrizing the surface as \mathbf{r} = z\cos{t}\mathbf{i} + z\sin{t}\mathbf{j} + z\mathbf{k} thus finding \iint_S\mathbf{F}\cdot\mathbf{n}\,dS = -\sqrt{2}\pi.
Obviously I wanted the result to be -\frac{\pi}{2} so that the total surface integral would be equal to zero which would verify the divergence theorem for this problem. But it's not and I can't figure out where I went wrong. Since -\sqrt{2}\pi is fairly close to -\frac{\pi}{2} I figure I'm on the right track at least with my working.
Hope this makes sense and thank you in advance for any assistance.