Lots of doubts abot the surface area of a cylinder.

Question

I recently started to study parametric surfaces, and I come across this exercise that I try to solve but I have a lot of doubts reganding the correctness of my resolution, and also I don't find similar examples on the internet.

I need to find the surface area of the cylinder $$x^{2} + y^{2} = 4x$$ bounded by z=0 and z+ x =4. The cylinder is centered at (2,0) with radius 2.

I made the parametrization $$<2+rcos(t) , rsen(t), 2-rcos(t)>$$ with r between 0 and 2, and t between 0 and 2π (First doubt : Is the parametrization right?)

Then, if everything is ok, I would proceed to do the formula of a surface area (I would not write the whole formula because I am very bad at MathJax). But you know, the double integral of the norm of the vector ("u") being "u" the cross product of the partial derivatives of the parametrization. The vector u in this case is (r,0,r) and the norm is $$\sqrt{2}r $$

Then, if everything is right, the area of the surface is the double integral

$$\int_0^{2π}\int_0^2\sqrt{2}*r^2 dθdr $$

Is this resolution right? If not, can you help me? Thanks.

PS : I know that the cylinder bounded by the plane is half of the full cylinder. This is the main reason that I think this resolution is wrong.

No, the parametrization isn't right. If you sketch the circle that generates the cylinder on the $;xy-,$ plane, it's easy to see that $;-\frac\pi2\le t\le \frac\pi2;$, for one....and where did you take the $;z-$ component from? — DonAntonio, Aug 06 '20 at 07:53
Hello. The plane that bounds the cylinder is z+x = 4. Its incorrect to say that z = x-4 then z = 2- rcos(t) ? — Ulises, Aug 06 '20 at 08:00
The answer is indeed the half of the area of a cylindre of radius $2$ and height $4$, $8\pi$. — , Aug 06 '20 at 08:07
@YvesDaoust Yes, I know that fact. And the resolution I made gives me other result, that is the reason of my doubts regarding my resolution. — Ulises, Aug 06 '20 at 08:10

score 3 · Accepted Answer · answered Aug 06 '20 at 08:16

3

Let us solve for

$$(x-2)^2+y^2=4,\\z\in[0,2-(x-2)].$$

in polar coordinates. We have

$$A=\int_0^{2\pi}\int_0^{2-2\cos\theta}\,dz\,2\,d\theta=2\int_0^{2\pi}(2-2\cos\theta)\,d\theta=8\pi$$ (the cosine term cancels out).

answered Aug 06 '20 at 08:16

I have a question : in the first double integral , the part "dz2dθ". Where does the number 2 comes from? Is the value of the radius? – Ulises Aug 06 '20 at 08:21
@Ulises I think for the Jacobian. – Sebastiano Aug 06 '20 at 08:21
@Ulises: yes, the element of area is $r,d\theta,dz$. – Aug 06 '20 at 08:22
Just a question. You can actually solve for $\theta$ as $\frac{\pi}{3}$. Is that significant or no? – C Squared Aug 06 '20 at 08:22
@Ulises: where did I change the order ? – Aug 06 '20 at 08:23
@YvesDaoust Then I have wrong? :-( Just a curiosity. Why $0\leq z \leq 1-(x-2)$? – Sebastiano Aug 06 '20 at 08:23
$z$ it the quote (height) of the cylinder. – Sebastiano Aug 06 '20 at 08:25
If you make the cylindrical substitutions for the given equation, you can get a value for $\theta$. – C Squared Aug 06 '20 at 08:25
We are using cylindrical coordinates. We have a circumference and z is the altitude. – Sebastiano Aug 06 '20 at 08:26
This one: $x^2+y^2=4x$ – C Squared Aug 06 '20 at 08:26
@Sebastiano: grr, what is given about $z$ ? – Aug 06 '20 at 08:27
By substituting $x=r\cos{\theta}$, $y=r\sin{\theta}$ – C Squared Aug 06 '20 at 08:28
If you simplify this you reach $\frac{1}{2}=\cos{\theta}$ – C Squared Aug 06 '20 at 08:29
I was just wondering if it was significant or not, as my original comment stated – C Squared Aug 06 '20 at 08:30
@CSquared: I meant expand your computation. – Aug 06 '20 at 08:31
$(x-2)^2+y^2=4(\cos{\theta}-1)^2+4\sin^2{\theta} = 4 $,,, $\cos^2{\theta}-2\cos{\theta}+1+\sin^2{\theta}= 3-2\cos{\theta}=1$ – C Squared Aug 06 '20 at 08:33
I thought $r=2$ – C Squared Aug 06 '20 at 08:36
Wow. I feel dummy stupid – C Squared Aug 06 '20 at 08:37
@CSquared: we are all subject to such confusions. But think that if both $r$ and $\theta$ are constant, you don't have a curve. – Aug 06 '20 at 08:38
That's why I was confused. Thanks for being understanding – C Squared Aug 06 '20 at 08:41
@CSquared: hope you didn't mind I let you understand by yourself. – Aug 06 '20 at 08:42
@YvesDaoust Sorry for asking again , but the 2 stands for the cross product of the vectors? – Ulises Aug 06 '20 at 08:51
@Ulises: vectors ? What vectors ? – Aug 06 '20 at 08:52
@YvesDaoust I ask because the formula in my original post (the double integral of the norm of the cross product blah blah). If you take the parametrization r(t,z) = (2+2cos(t), 2sen(t), z), the cross product of the partial derivatives of r is : (2cos(t), 2sen(t),0), and the norm of that is 2... – Ulises Aug 06 '20 at 09:45
@Ulises: to establish the element of area, you use normalized vectors. The $2$ comes from the Jacobian. – Aug 06 '20 at 09:49
You're right. I didn't even read the question right. It was in the hypothesis. I'm sorry, don't be angry. :-( – Sebastiano Aug 06 '20 at 10:32

DonAntonio · Answer 2 · 2020-08-06T08:18:53.510

First, you want "the skin" of the cylinder, not all what is there "in its interior", so the part to parametrize is only that: the envelope, and thus you can get

$$r(u,v)=(2+2\cos u,\,2\sin u,\,\text{something})\;,\;\;\text{with}\;\;-\frac\pi2\le u\le \frac\pi2$$

Now, that "something" must "advance vertically" from $\;z=0\;$ (which is the fancy name of the $\;xy-$ plane), and up until the plane $\;x+z=4\iff z=4-x=2-2\cos u\;$ (of course, it's not exactly "vertically" since there's a slant plane cutting the cyclinder. Also observe the plane and the circle in the $\;z=0\;$ plane intersect on this plane...)

Give it a try now.

Lots of doubts abot the surface area of a cylinder.

2 Answers2