Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

I have been asked to find a parameterization for the surface $9=x^2+z^2,0\leq y\leq4$, and rewrite the surface integral $\iint y dS$ as a double integral.

I believe that the parameterization should look like this.

$y=y,$ $ x=f(y)\cos\theta,$ $z=f(y)\sin\theta,$ $ 0\leq y\leq4 ,$ $0\leq \theta\leq2\pi $

I then should have $\textbf{r}(t)=\langle 3 \cos \theta,y,3\sin\theta\rangle$

However from this point on I'm not sure about how to convert it to arrive at the form $A(S)=\iint| \textbf{r}_u \times \textbf{r}_v|dA$

So building off Mhenni Benghorbal's answer I should arrive at $| \textbf{r}_u \times \textbf{r}_v| = \sqrt{9\cos^2v + 9\sin^2v } = 3$

Thus the surface integral is $\iint y dS = \iint u|\textbf{r}_u \times \textbf{r}_u|dA=\int^{2\pi}_0 \int^4_0 3v$ $du$ $dv = \int^{2\pi}_0 12dv = 24\pi$

share|cite|improve this question

Here is your parametrization of the surface $r(u,v)$ $$ r(u,v)= \langle 3 \cos v, u , 3\sin v \rangle, \quad 0 \leq u \leq 4,\, 0\leq v \leq 2\pi. $$

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.