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Checking divergence theorem inside a cylinder and under a paraboloid

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Problem 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

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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.
 
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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.
 
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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

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But


This 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.
 
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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

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

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

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