# Surface and Volume integration

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?

• Why do you think that it may be not correct ?
– Surb
Apr 8, 2020 at 14:51
• @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. Apr 8, 2020 at 14:53

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.