Use line integral to calculate the area of the surface that is the part of the cylinder defined by $x^2+y^2=4$, which is above the $x,y$ plane and under the plane $x+2y+z=6$.
I recently learnt that:
$\frac{1}{2}\oint\limits_{L}xdy – ydx\,=\frac{1}{2}\iint_D(1+1)=\text{Area of D}$. while $L$ is the curve around $D$. (Not sure if I translated it right).
And seems like I can't use that easily here, so
My understanding of the problem (First approach):
In this question, I guess I have to work with these steps (Some of them I'm not sure how to do them):
Find the intersections between the two planes and the cylinder and calculate their spaces using the formula above.
Find some equation that represents the sides of the wanted part of the cylinder? (This is the part I'm not sure about).
But a Problem might mess up my plan and that's if the planes intersect inside my cylinder, which will make the wanted part of cylinder split into two, and I have no idea how to deal with that.
Second approach:
Since there's a Cylinder then why not try to use Cylindrical coordinates (I am not fully $100 \%$ understanding why this might work, What I think of is we're transforming the xyz plane to $r,a,z$ plane that will be some much easier surface to calculate it's area):
$x=rcos(a), y=rsin(a), z=z, |J|=r$.
But here I am stuck again, I'm not sure how to use this in my line integral and jump to a double integral again, It's my first question of trying to calculate area using line integral and I'm finding it difficult to link between them.
I would appreciate any help, thanks in advance!
Best Answer
Parametrize the cylindrical surface as,
$r(\theta, z) = (2\cos\theta, 2\sin\theta, z)$
Now we know the surface area element of a cylinder is $dS = R \ dz \ d\theta = 2 \ dz \ d\theta$
or find $|r_{\theta} \times r_{z}| = 2$ (the radius of the cylinder)
$0 \leq \theta \leq 2\pi$. Lower bound of $z$ is $z = 0$ and upper bound is $z = 6 - x - 2y = 6 - 2 \cos\theta - 4 \sin\theta$
So surface area $S = \displaystyle \int_0^{2\pi} \int_0^{6 - 2 \cos\theta - 4 \sin\theta} 2 \ dz \ d\theta $
Edit (using line integral): please note that the question seeks surface area of only the cylinder's lateral surface.
So we define function $f(x,y) = 6 - x - 2y$ and we define our curve in XY-plane as $r (\theta) = (2 \cos \theta, 2 \sin \theta), 0 \leq \theta \leq 2\pi$
$|r'(\theta)| = 2$
$f(r(\theta)) = 6 - 2 \cos\theta - 4\sin\theta$
So the line integral to find surface area of the cylinder is,
$\displaystyle 2 \int_0^{2\pi} (6 - 2 \cos\theta - 4\sin\theta) \ d\theta$