I need help trying to use a line integral to find the lateral surface area of the part of a cylinder $x^2+y^2 =4 $ below the place $ x+2y+z =6 $ and above the $xy$- plane
I know how to find the surface area of a cylinder using line integrals but do i have to do this piecewise so it satisfies the other 2 conditions? Any help would be appreciated
Best Answer
Split the surface into pieces and calculate each individually then add:
For the bottom, we clearly have a circle of radius $2$ on the $xy$ plane. So the surface area there is $4\pi$.
For the surface area of top, we have a plane. That plane being $z=6-1x-2y$. Suppose we move in the $x$ direction one unit on the plane without moving in the $y$ direction. Clearly we move one unit down in the $z$ direction corresponding to the vector, $\langle 1,0,-1 \rangle$. Now suppose we move in the $y$ direction on unit this corresponds to the vector $\langle 0,1,-2 \rangle$. The cross product is $\langle 1,2,1 \rangle$. The magnitude of the cross product is $\sqrt{6}$. So we see that moving $1$ unit in strictly the $x$ direction then $1$ unit in strictly the $y$ direction on the $xy$ plane, we get an area of $\sqrt{6}$ on the plane $z=6-1x-2y$ for every unit square in the $xy$ plane.
Hence the area of the top is $\sqrt{6}(4\pi)$.
For the area of the side (cylinder), we need to evaluate,
$$\int_{C} z ds=\int_{C} (6-x-2y) ds$$
Where $C$ is the circle of radius $2$ in the $xy$ plane centered at the origin. Parametrize with $x=2\cos t$ and $y=2 \sin t$ then we have,
$$\int_{0}^{2\pi} (6-2\cos t-4 \sin t) dt$$
$$=12\pi$$
So the total surface area is,
$$12\pi+4\pi+(\sqrt{6}) 4\pi$$
$$=4\pi (3+1+\sqrt{6})$$
$$=\color{red}{4\pi (4+\sqrt{6})}$$