Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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:

$$\begin{align*} \iint_S \mathbf{A} \cdot {\boldsymbol{\hat{\textbf{n}}}} \,dS &= -\iint_D \mathbf{A}(\mathbf{u}(r,\theta)) \cdot \frac{\partial \mathbf{u}(r,\theta)}{\partial r} \times \frac{\partial \mathbf{u}(r,\theta)}{\partial \theta} \,dr \,d\theta \\&= -\int_{\theta=0}^{2\pi} \int_{r=0}^{2} (-2r^4\cos^3\theta-2r^4\sin^3\theta+r^5) \,dr \,d\theta \\&= -\left(0+0+\left[\theta\right]_{\theta=0}^{2\pi} \cdot \left[\frac{r^6}{6}\right]_{r=0}^{2}\right) = -\frac{64\pi}{3} \end{align*}$$

Note that the integral is negative; this should not come as a surprise, however, as the vector field and the normal vector point in opposite directions.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.