# Lots of doubts abot the surface area of a cylinder.

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 6 at 7:53
• 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 6 at 8:00
• The answer is indeed the half of the area of a cylindre of radius $2$ and height $4$, $8\pi$. – Yves Daoust Aug 6 at 8:07
• @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 6 at 8:10
• I know, I am confirming the value (that you didn't give). – Yves Daoust Aug 6 at 8:17

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).

• 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 6 at 8:21
• @Ulises I think for the Jacobian. – Sebastiano Aug 6 at 8:21
• @Ulises: yes, the element of area is $r\,d\theta\,dz$. – Yves Daoust Aug 6 at 8:22
• Just a question. You can actually solve for $\theta$ as $\frac{\pi}{3}$. Is that significant or no? – C Squared Aug 6 at 8:22
• @Ulises: where did I change the order ? – Yves Daoust Aug 6 at 8:23

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.