Finding the surface area of $G=\{(x,y,z):x^2+y^2\leq 1,0\leq z\leq x+1\}$

Question

Let $G=\{(x,y,z):x^2+y^2\leq 1,0\leq z\leq x+1\}$ find the surface area of $\partial G$

So it is a cylinder of radius $1$ bounded on the z-axis by $0$ and $x+1$. Can I say that because the maximum value of $x$ is $1$ then $z\in[0,2]$

So we have $\phi(u,r)=( \cos(u), \sin(u), r)$ so $\|\phi_u\times \phi_r\|=1$

So we have $\int_0^2\int_0^{2\pi}dudr=4\pi$?

Answer 1

First of all the surface isn't a "cylinder". In fact it's a part of a cylidner, where the upper and lower bound aren't parallel planes.

Anyway the idea to use polar coordinates is OK. For the side surface you can define the change of variable $(x,y,z) \to (\cos \theta, \sin \theta, z)$ and the magnitude of normal vector would be $1$. Hence the side surface is given by:

$$\int_0^{2\pi} \int_0^{\cos \theta + 1}dzd\theta = \int_0^{2\pi} (\cos \theta + 1) d\theta = \sin \theta + \theta \; \Bigg|_0^{2\pi} = 2\pi$$

Now you need to find the areas of the bottom and the top surface. The bottom surface is just a circle and the area would be $\pi$. On the other side the top can be parametrized by $(x,y,z) \to (r \cos \theta, r \sin \theta, r \cos \theta + 1)$. The magnitude of the normal vector would be $r\sqrt{2}$. Finally the surface is given by

$$\int_0^1\int_0^{2\pi}r\sqrt{2}d\theta dr = 2\pi \int_0^1 r\sqrt{2}dr = \sqrt{2}\pi$$

Finally adding all the values gives us that the total surface would be $(3 + \sqrt{2})\pi$

Answer 2

The shape is a cylinder of radius $1$ and height $2$, cut in half diagonally. The bottom is a circle of radius $1$, so that contributes area $\pi r^2 =\pi$. The lateral area of the whole cylinder is $2\pi r h = 4\pi$, so the lateral area of the cut cylinder is $2\pi.$ The top is an ellipse with major semi-axis $\sqrt{2}$ and minor semi-axis $1$. So its area is $\pi ab =\pi\sqrt{2}.$

All together that's $\pi\left(3+\sqrt{2}\right).$