# Find the outward flux of the vector field

Find the outward flux of the vector field $$F=(x^3,y^3,z^2)$$ across the surface of the region that is enclosed by the circular cylinder $$x^2+y^2=49$$ and the planes $$z=0$$ and $$z=2$$.

• What exactly is your problem? Do you know the how flux is calculated? Do you know what flux is? :) Show us your attempt. – u_sre Jul 4 '19 at 16:01

You can use the divergence theorem to evaluate the outward flux of the vector field.

Divergence theorem states the following: In other words we can simply add up the divergence in the region bound by our surface $$S$$, in order to calculate the outward flux of our vector field across our surface $$S$$.

You can read more here: https://en.wikipedia.org/wiki/Divergence_theorem

Now we can apply divergence theorem:

$$\nabla\cdot\mathbf{F}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}+\frac{\partial f}{\partial z}$$

$$\nabla\cdot\mathbf{F}=3x^2+3y^2+2z$$

Now we need to simply integrate over our region so we can evaluate: $$\int_0^2 \int_{-7}^7 \int_{-\sqrt{49-y^2}}^{\sqrt{49-y^2}} 3x^2+3y^2+2z \hspace{1mm} dx dy dz$$ But our integral is much easier if we use polar coordinates: $$\int_0^2 \int_{0}^{2 \pi} \int_{0}^{7} (3r^2+2z)r \hspace{1mm} dr d \theta dz$$

Evaluating this integral should get you $$7339\pi$$

• i got the same answer but its not correct – Ashish Paliwal Jul 4 '19 at 16:40
• I checked the answer with a calculator it should be correct. Did you copy down the question correctly? Are you sure the last term was $z^2$ and not $z^3$ – Anirudh Jul 4 '19 at 16:49