# Divergence of a vector field

Paraboloid R area is given by $z=x^2+y^2$ and limited by plate $z=4$

For the vector field $\vec A=x^2\vec i+y^2\vec j+z^2\vec k$

Find the value of $\iint\left(\vec A.\vec n\right) dS$

All the examples I solved there was a constant in paraboloid equation but here I can't figure out how to solve this.

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is it gonna be something like that $4=x^2+y^2$ ? – LuckyStrike Jan 6 '13 at 13:41
$4=x^2+y^2$ would describe a sphere, what you need to consider is a paraboloid with a flat cap. Try polar coordinates, the solution is not difficult once you get the geometry right. – anegligibleperson Jan 6 '13 at 13:59

Let $S$ be the surface of the paraboloid $z=x^2+y^2$ bounded by the circle $z=x^2+y^2=4$. An orientation for the surface has to be chosen, so let the normal vector point outward. A parametrization for the surface is given by $\mathbf{r}(x,y)=(x,y,x^2+y^2)$ for $x^2+y^2\le4$.
However, working in circular symmetry suggests that we use polar coordinates $(r,\theta)$. With $x=r\cos\theta$ and $y=r\sin\theta$ we obtain $z=x^2+y^2=r^2(\cos^2\theta+\sin^2\theta)=r^2$ and another parametrization $\mathbf{u}(r,\theta)=\mathbf{r}(r\cos\theta,r\sin\theta)=(r\cos\theta,r\sin\theta,r^2)$ for $r \in [0,2]$ and $\theta \in [0,2\pi)$. Let $D$ denote this domain in the $r\theta$-plane.
Let $\mathbf{A}$ be the vector field $\mathbf{A}(x,y,z)=(x^2,y^2,z^2).$ Note that $\mathbf{A}(\mathbf{u}(r,\theta))=(r^2\cos^2\theta,r^2\sin^2\theta,r^4)$. Now, since $\frac{\partial \mathbf{u}(r,\theta)}{\partial r}=(\cos\theta,\sin\theta,2r)$ and $\frac{\partial \mathbf{u}(r,\theta)}{\partial \theta}=(-r\sin\theta,r\cos\theta,0)$, it follows that \begin{align*} \frac{\partial \mathbf{u}(r,\theta)}{\partial r} \times \frac{\partial \mathbf{u}(r,\theta)}{\partial \theta}&=\left\lvert\matrix{ \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \cos\theta & \sin\theta & 2r \\ -r\sin\theta & r\cos\theta & 0 }\right\rvert\\&=(-2r^2\cos\theta,-2r^2\sin\theta,r(\cos^2\theta+\sin^2\theta))\\&=(-2r^2\cos\theta,-2r^2\sin\theta,r)\text{,} \end{align*} which is the normal vector pointing inward, so we have to change the sign in the surface integral. We can now evaluate it:
\begin{align*} \iint_S \mathbf{A} \cdot {\boldsymbol{\hat{\textbf{n}}}} \,dS &= -\iint_D \mathbf{A}(\mathbf{u}(r,\theta)) \cdot \frac{\partial \mathbf{u}(r,\theta)}{\partial r} \times \frac{\partial \mathbf{u}(r,\theta)}{\partial \theta} \,dr \,d\theta \\&= -\int_{\theta=0}^{2\pi} \int_{r=0}^{2} (-2r^4\cos^3\theta-2r^4\sin^3\theta+r^5) \,dr \,d\theta \\&= -\left(0+0+\left[\theta\right]_{\theta=0}^{2\pi} \cdot \left[\frac{r^6}{6}\right]_{r=0}^{2}\right) = -\frac{64\pi}{3} \end{align*}