Hey I have a question for this problem:
Let $U := (0, 4) × (0, 2\pi) ⊆ \mathbb{R}^2$ and $\varphi : U → \mathbb{R}^3$ given by $\varphi(r, θ) := (r \cos θ, r \sin θ, r)$.
Calculate $S_2(\varphi(U))$. So for $S_2(\varphi(U))$ I have this formula
$S_2(\varphi(U))=\int_U \det(D\varphi^t \cdot D\varphi)^{1/2}$
The problem is that here calculating $\det(D\varphi^t \cdot D\varphi)$ is really hard so I want to know if there is any trick to speed this up.
What I have done is:
$D\varphi=\left(\begin{matrix} \cos θ & -r\cos θ \\ \sin θ & r\cos θ \\ 1 & 0 \end{matrix}\right)$
And therefore $D\varphi^tD\varphi=\left|\begin{matrix} 2 & -r\cos^2θ+r\cos θ \sin θ \\ -r\cos^2θ+r\cos θ \sin θ & 2r^2\cos^2 θ \end{matrix}\right|$
And therefore $\det(D\varphi^t \cdot D\varphi)=\frac{1}{4}(6r^2+6r^2\cos 2θ+2r^2\sin 2θ+r^2\sin 4θ)$
It is really a lot of calculation and I wanted to know if there is some other way to "speed" this up and to make it "easier".
Doesn't look right. Your $(1,2)$ entry (top right) of $D\phi$ should be $-r\sin a$, so $(D\phi)^tD\phi=\begin{pmatrix}2 & 0\\ 0 & r^2\end{pmatrix}$.