The question asks to express the area of the surface as a double integral, using a parametrization then evaluate the integral. The portion of the plane y + 2z = 2 inside the cylinder x^2 + y^2 = 1.
I can't think of how to parametrize this tilted plane inside the cylinder, in general paramatrizing in R^3 isn't going so well for me, does anyone have any rules of thumb they use to come up with parametrizations in R^3.
Best Answer
Remember that parameterizing a surface means to describe each point $(x, y, z)$ on the surface using just two variables. In this case, we can write $z$ in terms of $y$: $z=1-\frac12y$
So our parameterization is $x = u, y = v, z = 1-\frac12v$. We can write this as a parametric vector function $\vec r(u, v) = \langle{u,v,1-\frac12v}\rangle$.
The restriction on parameters that we have is $u^2 + v^2 < 1$
The double integral expressing the surface area is $$\iint_R|\vec r_u\times\vec r_v|\ dA$$ $\vec r_u=\langle{1,0,0}\rangle$ and $\vec r_v=\langle{0,1,-\frac12}\rangle$, and you can get $|\vec r_u\times\vec r_v|$
In this case $R$ is just the area of the unit disk, which is $\pi r^2=\pi (1)^2=\pi$
Therefore the final answer would just be the result of $|\vec r_u\times\vec r_v|$ multiply by $\pi$.