Checking divergence theorem inside a cylinder and under a paraboloid

[JD_PM]

Homework Statement

$v = 9y\hat{i} + 9xy\hat{j} -6z\hat{k}$

Relevant Equations

$\int_{V}\nabla \cdot v dV = \int_{S} v \cdot da$

I am checking the divergence theorem for the vector field:

$$v = 9y\hat{i} + 9xy\hat{j} -6z\hat{k}$$

The region is inside the cylinder $x^2 + y^2 = 4$ and between $z = 0$ and $z = x^2 + y^2$

This is my set up for the integral of the derivative ($\nabla \cdot v$) over the region (volume); using cylindrical coordinates:

$$\int_{0}^{4}\int_{0}^{2\pi}\int_{0}^{2} (9rcos\theta - 6)r dr d \theta dz = -96\pi$$

Now the right hand side of the divergence theorem; the value of the function at the boundary(surface; i.e. its flux):

$$\iint_{R} v \cdot n \frac{dxdz}{|n \cdot j|} = 9\int_{0}^{4} \int_{-2}^{2} (x +x\sqrt{4-x^2})dxdz + 9\int_{0}^{4} \int_{-2}^{2} (-x +x\sqrt{4-x^2})dxdz$$

Is this later integral well arranged? $9\int_{0}^{4} \int_{-2}^{2} (x +x\sqrt{4-x^2})dxdz$ is vanishing and I don't get $-96\pi$ in the right hand side.

Thanks

 

[kuruman]

I don't understand your surface integral, specifically how you get the integrand $v \cdot n \frac{dxdz}{|n \cdot j|}$. What area element does this represent? A cylinder has three surface areas, two faces and a side, so there are three outward normals $\hat n$ to consider. Also, I would stick to cylindrical coordinates. That makes the expressions for $\hat n$ and the area elements much easier to write down.

Also, please use vector signs where they belong.

 

[JD_PM]

A cylinder has three surface areas, two faces and a side, so there are three outward normals $\hat n$ to consider.

Indeed, I have to compute the upper tap and the base fluxes. I will work out the RHS better.

 

[JD_PM]

But

This is my set up for the integral of the derivative ($\nabla \cdot v$) over the region (volume); using cylindrical coordinates:

$$\int_{0}^{4}\int_{0}^{2\pi}\int_{0}^{2} (9rcos\theta - 6)r dr d \theta dz = -96\pi$$

This is the LHS (the volume integral). Do you agree with the set up?

 

[LCKurtz]

ButThis is the LHS (the volume integral). Do you agree with the set up?

No. Those limits describe a solid cylinder. Your top is a paraboloid. You might find it more natural if your inside integral was the $dz$ integral.

 

[JD_PM]

No. Those limits describe a solid cylinder.

Absolutely, what a mistake sorry. I think this time you'll all agree. In Cartesian coordinates one gets:

$$\int_{-2}^{2}\int_{-\sqrt{4-x^2}}^{-\sqrt{4-x^2}}\int_{0}^{x^2 + y^2} (9x - 6) dz dx dy$$

But it is nicer to work with cylindrical coordinates:

$$\int_{0}^{2\pi}\int_{0}^{2}\int_{0}^{r^2} (9rcos \theta - 6)rdz dr d\theta$$

 

[LCKurtz]

Absolutely, what a mistake sorry. I think this time you'll all agree. In Cartesian coordinates one gets:

$$\int_{-2}^{2}\int_{-\sqrt{4-x^2}}^{-\sqrt{4-x^2}}\int_{0}^{x^2 + y^2} (9x - 6) dz dx dy$$

But it is nicer to work with cylindrical coordinates:

$$\int_{0}^{2\pi}\int_{0}^{2}\int_{0}^{r^2} (9rcos \theta - 6)rdz dr d\theta$$

Not only nicer, but correct. You have a couple of errors in the rectangular version.

 

[JD_PM]

Not only nicer, but correct. You have a couple of errors in the rectangular version.

True:

$$\int_{-2}^{2}\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\int_{0}^{x^2 + y^2} (9x - 6) dz dy dx$$

 

[JD_PM]

$$\int_{-2}^{2}\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\int_{0}^{x^2 + y^2} (9x - 6) dz dy dx = -48 \pi$$