Find surface area of cylinder $x^2+z^2=1$ between two planes $x+y-4=0, y-z+4=0$ Find surface area of cylinder $x^2+z^2=1$ between two planes $x+y-4=0, y-z+4=0$.
I drew the picture but I can't seem to figure out the way to start.. If someone could give me a hint on how to start it would mean a lot..
 A: Given your cylinder is along $y-$axis and radius is $1$, your parametrization should be
$x = \cos t, y, z = \sin t$
$z - 4 \leq y \leq 4 - x$
Also $0 \leq t \leq 2\pi$
Now $|r'_y \times r'_t| = r = 1$ (in this case). You can check that.
Your integral to find surface area is $\iint |r_y \times r_t| \, dy \, dt$
A: Projects everything on the yz plane and works as a solid of revolution. I think it is the first idea that comes to mind, I hope it helps you
A: If you rotate planes about the y axis to be parallel then the cylinder has a constant height of 8.  The sides are then $2 \pi r h = 16 \pi$.  The planes intersect at a $45^\circ$ angle so the top and bottom are $r1 \cdot r2 \cdot \pi = 1 \sqrt{2} \ pi$.
$$A = (16+2\sqrt{2}) \pi$$
A: With cylindrical coordinates $x= r\cos \theta$ and $z=r \sin\theta$, the surface between the two planes can be integrated along the cross-section circle of radius $r=1$
$$A= \int_0^{2\pi} (y_2-y_1)\cdot 1\cdot d\theta=
 \int_0^{2\pi} ((4-\cos\theta )-(\sin\theta -4))d\theta=16\pi
$$
