surface of $M=\{(x,y,z)\in\mathbb R^3: x^2+y^2<1 \text{ and } z=3-2x-2y\}$

Question

For $M=\{(x,y,z)\in\mathbb R^3: x^2+y^2<1 \text{ and } z=3-2x-2y\}$ I want to determine the surface of $M$, this means I want to calculate $\int_M 1 dS_M$. I don't know how to this and I only need a hint how to start. I think first of all I need a diffeomorphism $\varphi : U\rightarrow M$, for a open set $U$. Can I chose polar coordinates? Thanks

Answer 1

You have a cylinder in the definition of $M$. So it's better to use cylindrical coordinates. Also, notice that you must find the area of a plane that is restricted to a cylinder, i.e. the area of an ellipse, that equals the product of minor and major radiuses times $\pi$. So let $(x, y, z) = (r\cos\theta, r\sin\theta, z)$, then $$M = \{(r, \theta, z) \in \mathbb{R}^{\geq 0} \times [0, 2\pi] \times \mathbb{R} : r < 1 \mbox{ and } z = 3 - 2r(\cos\theta + \sin\theta)\}.$$ So the equation of the ellipse is $z = 3 - 2(\cos\theta + \sin\theta)$. It's obvious that the minor radius is $R_{\text{minor}} = 1$ (because the raduis of the cylinder is $1$). To find the major radius, we must find the maximum and minimum of $z$ in $z = 3 - 2(\cos\theta + \sin\theta)$. Thus \begin{align*} \cos\theta + \sin\theta = \sqrt{2}\sin(\theta + \pi/4) \implies \left\{ \begin{array}{ll} \mbox{maximum} = z_M = \sqrt{2} &\mbox{ for }\ \theta_M = \pi/4\\ \mbox{minimum} = z_m = -\sqrt{2} &\mbox{ for }\ \theta_m = 5\pi/4 \end{array} \right. \end{align*} So to find the major radius, we must calculate the distance of $(1 \times \cos\theta_M, 1 \times \sin\theta_M, z_M)$ and $(1 \times \cos\theta_m, 1 \times \sin\theta_m, z_m)$ divided by $2$, which equals $R_{\text{major}} = 3$. Therefore we have: $$\int_M 1\ dS_M = \pi R_{\text{major}}R_{\text{minor}} = \pi \times 3 \times 1 = 3\pi$$