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Given a set $M = \{(s,t,u) \mid 0 < s < t < 2\pi, 0 < u < 2\} \subseteq \mathbb{R}^3$ and

\begin{align} f: M \to \mathbb{R}^3, \quad f(s,t,u) = (us\cos t, us \sin t, us + ut) \end{align} with $f(M) := G = \{ f(s,t,u) \mid 0 < s < t < 2\pi, 0 < u < 2\} \subseteq \mathbb{R}^3$ and $S = \{f(s,t,1) \mid 0 < s < t < 2 \pi\} \subseteq G$.

I would like to compute the surface of $S$ and the volume of $G$ but I am unsure whether my idea is correct or not...

I attempted to use the transformation law. Hence, I compute the Jacobian $J$ of $f$:

\begin{pmatrix} u \cos t & -us \sin t & s \cos t \\ u \sin t & us\cos t & s \sin t \\ u & u & s+t \end{pmatrix} with $\det J = u^2t \neq 0$.

The surface of $S$ is given by

\begin{align} \int_0^{2\pi} \int_0^t u^2 \cdot t \ \mathrm{d}s \ \mathrm{d}t = \int_0^{2\pi}\left[ u^2ts\right]_0^t \mathrm{d}t = \left[\frac{u^2t^3}{3}\right]_0^{2\pi} = \frac{8\pi^3}{3}u^2. \end{align}

and the volume of $G$ would be given by

\begin{align} \int_0^{2\pi} \int_0^t \int_0^2 u^2 \cdot t \ \mathrm{d}u \ \mathrm{d}s \ \mathrm{d}t = \int_0^{2\pi} \int_0^t \frac{8}{3} \cdot t \ \mathrm{d}s \ \mathrm{d}t = \int_0^{2\pi}\left[ \frac{8}{3} ts\right]_0^t \mathrm{d}t = \left[\frac{\frac{8}{3} t^3}{3}\right]_0^{2\pi} = \frac{8\pi^3}{3}\frac{8}{3} = \frac{64}{9}\cdot \pi \end{align}

Am I correct?

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  • $\begingroup$ Why do you think that it may be not correct ? $\endgroup$
    – Surb
    Apr 8, 2020 at 14:51
  • $\begingroup$ @Surb Lack of confidence. The computations should be fine, it's more that I'm unsure about whether or not I did something systematically wrong. $\endgroup$
    – offline
    Apr 8, 2020 at 14:53

1 Answer 1

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Your idea for surface area is incorrect and your approach for volume is correct. Recall a fact that to write the parametric equation of surface we need two independent variables. $S = \{f(s,t,1) \mid 0 < s < t < 2 \pi\}= (s\cos t, s\sin t, s+t) $ so you can write parametric equation of surface $S$ in vector form $\vec r(s,t)=s\cos t \, \hat i +s\sin t \, \hat j + (s+t) \hat k $ now surface area of s is given by $$\int_0^{2\pi} \int_0^t \ |\vec r_s \times \vec r_t| \ \mathrm{d}s \ \mathrm{d}t $$ where $\vec r_s = \frac {\partial \vec r}{\partial s} = ( \cos t , \sin t , 1)$ and $\vec r_t = \frac {\partial \vec r}{\partial t} = ( -s \sin t , s\cos t , 1)$ now calculate $\vec r_s \times \vec r_t = (\sin t - s\cos t , -s\sin t, -\cos t)$ which is simply cross product of these two vectors. Now you can do calculations.

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