Volume integral, am I doing this right?

[Damascus Road]

Hi all,

evaluating $$\int\int\int \nabla . V d\tau$$ over

$$x^{2} + y^{2} \leq 6, 0 \leq z \leq 10$$

where V is a vector function of just $$\hat{x}$$ and $$\hat{y}$$.

Using the divergence theorem, and doing the dot product of V with the normal of the first surface,

the two partials w.r.t x and y are 2x and 2y,
doing the cross product, the normal is 4xy z.

So, the dot product is zero since there is no z component to V?

Then, when I go to do the same for the second surface, I'm not sure what my double integral limits are, since I only have a surface of z...

help!
thanks!

 

[mighty2000]

Hello,

Thank you for your post. I understand your confusion regarding the limits of the double integral for the second surface. Let me try to clarify it for you.

Firstly, you are correct in your understanding that the dot product of V with the normal of the first surface is zero since there is no z component to V. This is because the vector function V only has components in the x and y directions.

Now, for the second surface, the limits of the double integral will depend on the specific shape of the surface. However, since the surface is defined by z=0, the limits of the double integral will be from x=0 to x=√(6-y^2) and from y=0 to y=√6. This is because the surface is a circle in the xy-plane with radius √6. Therefore, the double integral will be evaluated over this circular surface.

I hope this helps clarify things for you. If you have any further questions, please feel free to ask. Thank you.