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Find surface area of cylinder

  1. May 16, 2012 #1

    sharks

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    1. The problem statement, all variables and given/known data
    Find the surface area of the portion of the cylinder [itex]x^2+y^2=4y[/itex] lying inside the sphere [itex]x^2+y^2+z^2=16[/itex]


    2. Relevant equations
    Double integral for finding surface area.
    [itex]\iint_S d\sigma[/itex]

    3. The attempt at a solution
    At first, i thought it was just the area of the surface of sphere found inside the cylinder which goes up and down infinitely and is parallel to the z axis. But then it occurred to me that it was about finding the lateral surface area of the cylinder, which is a tad more complicated.

    See attached graph for cylinder projection on the x-y plane. The sphere has a radius of 4, so it completely encloses the cylinder in the x-y plane, until the latter shoots through the sphere.

    Here is my plan to find the lateral area of the cylinder: split the cylinder in half across y-z plane and then project one half on the y-z plane and then finally multiply the answer by 2 to get the whole surface area.

    But if i substitute the equation of the cylinder into the equation of the sphere, i get [itex]4y+z^2=16[/itex] which gives me the equation of points of intersection at the surface between the sphere and cylinder. See second attached graph. I can use that equation to find the limits later on.

    So, considering the equation of the cylinder directly. Since i will be projecting onto the y-z plane: [itex]x=\sqrt{4y-y^2}[/itex]. Then, i find ##x_y## and ##x_z## and the usual steps to finding the surface area gives:
    [tex]\iint_Ω \frac{2}{\sqrt{4y-y^2}}\,.dydz[/tex]
    [tex]\int^{z=4}_{z=-4} \int^{y=\frac{16-z^2}{4}}_{y=0} \frac{2}{\sqrt{4y-y^2}}\,.dydz[/tex]
    I'm stuck with the integral.
    I tried with polar coordinates but it just gets more complicated. Any advice?
     

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    Last edited: May 16, 2012
  2. jcsd
  3. May 16, 2012 #2

    LCKurtz

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    That is an interesting, but not easy, problem. I would parameterize it in cylindrical coordinates to begin with. The first thing to show is that the equation of the cylinder in cylindrical coordinates is ##r = 4\sin\theta## for ##0 \le \theta \le \pi##. (You can do that, right?) So parameterizing the cylinder gives$$
    \vec R(\theta,z)=\langle r\cos\theta, r\sin\theta,z\rangle=\langle 4\sin\theta\cos\theta,4\sin^2\theta,z\rangle$$Now, as you have noticed, it is tricky figuring out the ##z## limits, which will depend on ##\theta##. The sphere in cylindrical coordinates is ##r^2+z^2=16##, so put ##r=4\sin\theta## in that to get the relation between ##z## and ##\theta## on the intersection curve. You should get ##z=\sqrt{16-16\sin^2\theta }= 4|\cos\theta|##.

    You now have the parameterization and the limits. Do you know how to figure out dS and set up the integral from here?
     
    Last edited: May 16, 2012
  4. May 16, 2012 #3

    sharks

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    Hi LCKurtz
    This doesn't appear right. The equation of the cylinder in polar form:
    [tex]r^2=4r\sin \theta
    \\r^2-4r\sin \theta=0
    \\r=4\sin \theta[/tex]
     
  5. May 16, 2012 #4

    LCKurtz

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    Yes, missed that typo when I corrected my first post. ##r=4\sin\theta## is right, but you want ##\theta## from ##0## to ##\pi## to keep ##r## positive, which is standard.

    I'm going to re-edit my post with the ##4##.
     
  6. May 16, 2012 #5

    sharks

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    Finding the surface area of the projection of the cylinder onto the y-z plane. From the equation of cylinder:
    [tex]x=\sqrt{4y-y^2}[/tex]I found ##x_y## and ##x_z##
    [tex]S.A.=\iint_\Omega \sqrt{(x_y)^2+(x_z)^2+1}\,.dydz
    \\=\iint_\Omega \frac{2}{\sqrt{4y-y^2}}\,.dydz[/tex]which is what i've done in post #1. Now, i just have to parametrize it, and replace y by [itex]r\sin \theta[/itex] and replace dydz by [itex]rdrd\theta[/itex].
    But this is the same procedure as what i did in my post #1. I'll just end up with the same integral as in post #1 except in polar coordinates.

    LCKurtz, i'm not sure i understand your procedure. Is your suggestion about finding the lateral surface area of the section of cylinder within the sphere? Or the surface area of sphere inside the cylinder?

    The cylinder is viewed as the right-hemisphere when projected onto the y-z plane. The total lateral surface area of cylinder is:
    [tex]2\int^{\theta =\pi}_{\theta =0} \int^{r=4}_{r=0} \frac{2}{\sqrt{4r\sin \theta-r^2\sin ^2 \theta}}\,.rdrd\theta[/tex]
    Now, i'm not sure how to solve the integral.
     
    Last edited: May 16, 2012
  7. May 16, 2012 #6
    Because of my bad English, I will show how to do this problem:

    Find the surface area of the portion of the cylinder [itex]x^2+z^2=b^2[/itex] lying inside the sphere [itex]x^2+y^2+z^2=a^2[/itex] where [itex]0<b<a[/itex].

    Solution: [itex]P=2P(S_1)=2\iint_{S_1}^{}\mbox{d}S=2\iint_{D}^{} \sqrt{1+(y_x)^2+(y_z)^2}\mbox{d}x\mbox{d}z[/itex] where [itex]S_1:~y=+\sqrt{a^2-x^2-z^2}[/itex] and [itex](x,z)\in D, \quad D:~x^2+z^2\le b^2[/itex].

    This is a sketch.
     
  8. May 16, 2012 #7

    LCKurtz

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    Yes. I have shown you the parameterization in terms of ##z## and ##\theta## and the appropriate limits. So I will ask you again: Do you know the general formula for ##dS## for a surface parameterized as ##\vec R = \vec R(u,v)##?
     
  9. May 16, 2012 #8

    sharks

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    I would say that dS is the differential area, but since i'm not sure what you meant, i will go with "no". In fact, i'm not sure why you had to parametrize in terms of ##z## and ##\theta##
     
    Last edited: May 16, 2012
  10. May 16, 2012 #9

    LCKurtz

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    You don't have to use ##z## and ##\theta##; you can use any parameters you wish. But they are the natural parameters to use for this cylinder. ##4\sin\theta## locates you on the circle and ##z## tells you how far up the cylinder you are.

    If you haven't studied parametric surfaces and their areas, you wouldn't understand what I have shown you. But if you have you should have seen the formula for ##\vec R = \vec R(u,v)##$$
    dS = |\vec R_u \times \vec R_v|\, du dv$$Have you actually not studied that yet?
     
  11. May 16, 2012 #10

    sharks

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    I studied linear algebra, calculus 1 & 2 (according to my university syllabus) by myself from whatever i could salvage. I might not have seen the same notations, but i think i've covered all of the essentials.

    Are you referring to the Jacobian? It looks like the transformations from dxdy into dudv.
     
  12. May 16, 2012 #11

    LCKurtz

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    Well, I can't give you a course in it here. To finish the problem as I have started you have to calculate$$
    dS = |\vec R_\theta \times \vec R_z|\, dzd\theta$$You will be surprised how simple that works out to be. Then work out the integral$$

    Area = \int_0^\pi \int_0^{4|\cos\theta|}|\vec R_\theta \times \vec R_z|\, dzd\theta$$I got an answer of 32, assuming I didn't make any mistakes. Here's a picture (the green is the portion of the cylinder for the area calculation):
    cylinderinsphere.jpg
     
  13. May 16, 2012 #12

    sharks

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    My understanding for finding the surface area is more directly related to what Karamata has posted.

    But the method you have suggested using [itex]dS = |\vec R_\theta \times \vec R_z|\, dzd\theta[/itex] is that the only way to solve this particular problem? Or can it be solved using what i already know? Just asking.

    I'll pick up some books in the library tomorrow and will investigate more on your suggestion.
     
  14. May 18, 2012 #13

    sharks

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    OK, i managed to work it out, but using a somewhat different method that was hard to find, as it was only a small paragraph in my notes.

    The lateral surface area of cylinder is obtained from the line integral:
    [tex]\int_C (z_2-z_1)\,.ds[/tex] where C is the equation of the circle which forms the base of the cylinder in the x-y plane and ##z_2## is the upper surface and ##z_1## is the lower surface of the cylinder.
    Using the same parametrized equations derived in the posts above; [itex]r=4\sin \theta[/itex] and ##z=4\cos \theta##
    [tex]\frac{dx}{d\theta}=4\cos 2\theta,\;\frac{dy}{d\theta}=4\sin 2\theta[/tex]
    [tex]z_1=0,\;z_2=4\cos \theta[/tex]
    [tex]ds=\sqrt{ \left( \frac{dx}{d\theta}\right)^2 + \left( \frac{dy}{d\theta}\right)^2}\,.d\theta=4d\theta[/tex]
    So, the line integral becomes:
    [tex]\int^{\pi /2}_0 4\cos \theta \,.4d\theta=16[/tex]
    To get the whole surface area, i multiply by 4, and the answer is 64.

    I'm confused about the limits of the integral, as i initially wanted to set it from 0 to ∏, but the evaluation of the integral gives 0, so i had no choice but to cut down the limit in order to obtain a value. As a general check, suppose that i integrate over a circular region that touches the origin, in polar coordinates, does that mean that the total integration will always be zero?
     
  15. May 21, 2012 #14

    sharks

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    I learned your suggestion. I realize that i had previously only done explicit formula for surface area. This problem relates to parametrization of surface area, which i just read about, and there is a third formula which deals with implicit surface area, which i have not done, and probably not included in my syllabus.

    [tex]\vec R(\theta,z)=\langle r\cos\theta, r\sin\theta,z\rangle=\langle 4\sin\theta\cos\theta,4\sin^2\theta,z\rangle
    \\\vec R(\theta,z)= 4\sin\theta\cos\theta\hat i+4\sin^2\theta\hat j+z\hat k=2\sin 2\theta\hat i+4\sin^2\theta\hat j+z\hat k
    \\\vec R_\theta=4\cos 2\theta\hat i+8\sin\theta\cos\theta\hat j=4\cos 2\theta\hat i+4\sin 2\theta\hat j
    \\\vec R_z=\hat k
    \\\vec R_\theta \times \vec R_z=\begin{vmatrix}\hat i & \hat j & \hat k \\ 4\cos 2\theta & 4\sin 2\theta & 0 \\ 0 & 0 & 1\end{vmatrix}=(4\sin 2\theta)\hat i - (4\cos 2\theta)\hat j+0 \hat k
    \\|\vec R_\theta \times \vec R_z|=\sqrt{(4\sin 2\theta)^2 + (4\cos 2\theta)^2}=4
    \\dS = |\vec R_\theta \times \vec R_z|\, dzd\theta=4\,dzd\theta
    \\\int_0^\pi \int_0^{4|\cos\theta|}4\,dzd\theta=0[/tex]I don't understand how you got the area of 32.

    But if i split the region above the plane z = 0 in half across the plane y = 0, then the limits become ##0\le \theta \le \pi /2 ## and the double integral becomes:
    [tex]2\int_0^{\pi /2} \int_0^{4|\cos\theta|}4\,dzd\theta=32[/tex]
    The method that i've used in the above post #13 is actually part of my notes, but in the sole given example, it relates to a cylinder with its vertical axis as the z axis and bounded between 2 planes ##z_2## and ##z_1##, but this problem here is different in 2 ways:
    1. This cylinder's axis is not the z axis.
    2. This cylinder is bounded between a smooth surface and a plane.
    Although i struggled for a long time and repeatedly to understand what i might be doing wrong, i think the line integral formula that i used is limited as it might only relate to cylinders which are bounded between planes, and not surfaces. Or maybe the line integral formula doesn't work either if the cylinder's axis is not the z axis, but the latter isn't convincing.
    To conclude, i think that the line integral formula for calculating the lateral surface area of cylinder, relates only to planes, just like the Green's theorem, hence why the formula in my notes didn't work on this problem. Do you happen to know anything about that line integral method?

    On a side note, how did you plot that graph? What software did you use? I tried using various graph plotters, like wolfram, etc, but couldn't get the same result.
     
    Last edited: May 21, 2012
  16. May 21, 2012 #15
    Before I start, I'm going to say that my knowledge of calculus is limited, and I have never worked with the three-dimensional axis (I'm a graduating high school senior in calc BC). And, unfortunately, I haven't yet been able to figure out this fancy latex form of typing.

    But the way I would have approached this problem (if given the image in post #11) would be to accumulate the heights of the cylinder around half of the body in the +z area of space, and quadrupole my result.

    If using theta from 0 to pi, the x coordinate would be 2sin(theta) and the y coordinate would be 2-2cos(theta), meaning the height would be √{16 - 4sin2(theta) - 4 + 8cos(theta) - 4cos2(theta)} = 2√{2 - 2cos(theta)}.
    Then the answer would be 8∫√{2 - 2cos(theta)} from 0 to pi.

    Evaluated on my calculator, I get 32 as an answer, but I can't see a way to integrate this by hand. (Probably why the more advanced methods discussed here are necessary :b)
     
  17. May 25, 2012 #16

    LCKurtz

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    Hello Sharks. I haven't been purposely ignoring you. Real life intervened and I spent several days in the hospital for the first time in my life. Anyway, to answer your question, I plotted that picture with Maple.
     
  18. May 25, 2012 #17

    sharks

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    Hi LCKurtz!

    I really hope that you are feeling better now. Personally, i have the greatest admiration and respect for the PF Homework Helpers and mentors. I want to thank you for all that you do here. Initially, i didn't want to bother you with a PM (as i know that you have a lot of responsibilities already) but i noticed that you weren't online for some time, which is quite unusual for you, given your obvious passion for mathematics and daily involvement in PF. I wish you a quick and full recovery.

    Your help, advice and patience is greatly appreciated. But real life always comes first, and you should take some time off and relax, every now and then. You definitely deserve it.

    Regards,
    S
     
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