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Find area of surface of sphere

  1. Jan 2, 2012 #1

    sharks

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    The problem statement, all variables and given/known data
    http://s1.ipicture.ru/uploads/20120102/3slDHh7g.jpg

    The attempt at a solution
    I drew the graph in my copybook. The sphere has radius a from the center (0,0)

    The cylinder has center (a/2,0) and radius a/2

    Then, i don't know what's the next step.
     
  2. jcsd
  3. Jan 2, 2012 #2
    The formula for the surface area of [itex]z = f(x, y)[/itex] above some region [itex]D[/itex] is:

    [tex] SA=\underset{D}{{\iint}} \sqrt{1+(f_x)^2+(f_y)^2}\: dA[/tex]

    (Alternatively, write down a surface integral and parametrize the sphere - but it will end up being the same thing.) The question now is: what is the [itex]f(x, y)[/itex] to use, and what is [itex]D[/itex]?
     
  4. Jan 3, 2012 #3

    sharks

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    OK, i'll make an attempt:

    Since [itex]z = f(x, y)[/itex], then [itex]z=√(a^2-x^2-y^2)[/itex]

    [itex]D[/itex] is the equation of the cylinder. On the x,y,z Cartesian plane, [itex]y=√(ax-x^2)[/itex]

    ... and then i don't know.
     
  5. Jan 3, 2012 #4

    ehild

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    Think of an apple corer, used not through the centre of the apple.

    pomme1.jpg
    You integrate for a domain is the x,y plane. How this domain looks like? What are the bounds for x and y?

    ehild
     
  6. Jan 3, 2012 #5

    sharks

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    ehild, now i get a clearer mental picture of how this problem exists in 3-D. But from what my tutor explained in the previous chapters relating to double and triple integral, i must always first describe the region and then use those boundaries to write the limits. This problem is one of the first in the section "Surfaces" in my notes, but the explanations are scarce, so i'm struggling a little to find my way around.

    I will first try to describe the region:

    For x and y fixed, z varies from [itex]z=0[/itex] to [itex]z=√(a^2-x^2-y^2)[/itex]
    For x fixed, y varies from [itex]y=0[/itex] to [itex]y=√(ax-x^2)[/itex]
    x varies from [itex]x=0[/itex] to [itex]x=a[/itex]

    OK, here is another attempt (based loosely on an example in my notes, which i'm not sure applies to this problem here):

    Since the sides of the cylinder is parallel to the z-axis, a projection on the xy-plane is not possible. We must therefore project the area onto the xz or yz plane. So, i've decided to project on the xz-plane. The projection is a triangle AOB. The equation of the surface will be of the form y= g(x,z).
    Then, i use this formula [itex]\sec \beta=\underset{D}{{\iint}} \sqrt{1+(f_x)^2+(f_z)^2}\: dA[/itex]
    But [tex]f_x=\frac{\partial y}{\partial x}[/tex] and [tex]f_z=\frac{\partial y}{\partial z}[/tex]
    From equation of sphere, [itex]y=\sqrt(a^2-x^2-z^2)[/itex]. Therefore, [itex]f_x=\frac{-x}{\sqrt(a^2-x^2-z^2)}[/itex] and [itex]f_z=\frac{-z}{\sqrt(a^2-x^2-z^2)}[/itex]

    So, i describe the region again, before writing the limits of the double integral:
    Since the projected area is found only in the xz-plane, then the y-axis is ignored.

    For x fixed, z varies from [itex]z=0 \; to\; z=√(a^2-x^2-y^2)[/itex]
    x varies from [itex]x=0\; to\; x=a[/itex]

    Hence, the surface area, [itex]\sigma=\int\int\sec \beta\,.dzdx[/itex]

    I'm not sure that all i typed above is correct or even relevant to this problem. From the picture above, i think that the surface area needed is the surface area of only the apple skin of the cut section. Correct?
     
    Last edited: Jan 3, 2012
  7. Jan 3, 2012 #6

    ehild

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    Why not? Is not the projection of the cylinder on the (x,y ) plane a circle of radius a/2 and centre at (a/2, 0)?
    You can project the thing on any plane. See the projections on the xz and xy planes in the attachment.

    Think the apple corer is like your cylinder, its axis parallel to the z axis and imagine the piece of apple when you remove it from the cylinder of the corer.

    ehild
     

    Attached Files:

    Last edited: Jan 3, 2012
  8. Jan 3, 2012 #7

    sharks

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    OK, i think i understand. Your graph was very helpful. So, the surface area that i need to find is the surface area of the apple skin removed from the "piece of apple removed from the corer". That would mean the top surface (of the sphere) and bottom surface (of the sphere). Two pieces of skin. Correct?

    In that case, what do i modify in my calculations from my previous reply above? I'm confused.
     
    Last edited: Jan 3, 2012
  9. Jan 3, 2012 #8

    ehild

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    Yes, you need to determine the area of both pieces of skin.
    But you can not ignore the cylinder from the boundary of the region.

    ehild
     
  10. Jan 3, 2012 #9

    sharks

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    So, i can find the surface area of only the top or bottom part and then multiply by 2 to get the total surface area.

    Description of the top region:
    For x and y fixed, z varies from [itex]z=0[/itex] to [itex]z=√(a^2-x^2-y^2)[/itex]
    For x fixed, y varies from [itex]y=0[/itex] to [itex]y=√(ax-x^2)[/itex]
    x varies from [itex]x=0[/itex] to [itex]x=a[/itex]

    I can transform the Cartesian coordinates to Polar coordinates, in an attempt to simplify the integration. But i don't think that it's correct, as this method would give me the volume, instead of the surface area.

    I think i should go back to this explanation:
    From [itex]z=√(a^2-x^2-y^2), f_x=\frac{-x}{\sqrt(a^2-x^2-z^2)}[/itex] and [itex]f_y=\frac{-y}{\sqrt(a^2-x^2-z^2)}[/itex]

    The boundaries for region D:
    For x fixed, y varies from [itex]y=0[/itex] to [itex]y=√(ax-x^2)[/itex]
    x varies from [itex]x=0[/itex] to [itex]x=a[/itex]

    Is that correct?
     
    Last edited: Jan 3, 2012
  11. Jan 3, 2012 #10

    ehild

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    Take care, f is function of x and y. Change z to y in the derivatives fx and fy. y varies from -√(ax-x2) to √(ax-x2), but you can integrate from y=0, and then multiply the integral by two.
    But the integration looks very ugly.
    ehild
     
    Last edited: Jan 3, 2012
  12. Jan 3, 2012 #11

    sharks

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    So, [itex]f_x=\frac{-x}{y}[/itex] or [itex]f_x=\frac{x}{y}[/itex]

    and [itex]f_y=1[/itex] or [itex]f_y=-1[/itex]

    Now, i should probably convert to polar coordinates to integrate?

    Using x=rcosθ, y=rsinθ and [itex]x^2 + y^2 = r^2[/itex]

    Converting the integrand to polar coordinates: [itex]\sqrt{cosec^2 \theta + 1}[/itex]

    The new limits:
    For θ fixed, r varies from r=0 to r=acosθ
    θ varies from θ=0 to θ=∏/2 (this only gives me a semi-circle, so the final answer should be multiplied by 2? or i could also say, additionally θ varies from θ= -∏/2 to θ=0?)

    I am having trouble evaluating this last integral:
    [tex]\sigma = 2\int_0^\frac{\pi}{2} acos\theta \sqrt{cosec^2 \theta + 1}\,.d\theta[/tex]
     
    Last edited: Jan 3, 2012
  13. Jan 3, 2012 #12

    ehild

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    Remember, it is still the surface of the sphere.
    [tex]f(x,y)=z=\sqrt{a^2-x^2-y^2}[/tex]

    You did the derivations correctly before.
    [tex]f_x=\frac{-x}{\sqrt{a^2-x^2-y^2}}[/tex]
    [tex]f_y=\frac{-y}{\sqrt{a^2-x^2-y^2}}[/tex]

    Your limits are correct, integrate from zero to pi/2 with respect to theta, and multiply by two.
     
  14. Jan 3, 2012 #13

    sharks

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    So, transforming the integrand to polar coordinates:

    [tex]f_x=\frac{-r\cos\theta}{\sqrt{a^2-r^2}}[/tex]
    [tex]f_y=\frac{-r\sin\theta}{\sqrt{a^2-r^2}}[/tex]
    [tex]SA=\underset{D}{{\iint}} \sqrt{1+(f_x)^2+(f_y)^2}\: dA=\int_0^\frac{\pi}{2}\int_0^{a\cos\theta} \sqrt{\frac{a^2}{a^2-r^2}}\: rdrd\theta[/tex]
    If that integrand is correct, then i'm not sure how to proceed with its integration. It looks complicated, but i will make an attempt:
    [tex]SA=\int_0^\frac{\pi}{2}\int_0^{a\cos\theta} ar\frac{1}{\sqrt{a^2-r^2}}\: drd\theta=\int_0^\frac{\pi}{2} -a\sqrt{a^2-r^2}\,.d\theta=\int_0^\frac{\pi}{2} a^2-a^2 \sin\theta\,. d\theta=a^2\theta+a^2\cos\theta=a^2(\frac{\pi}{2}-1)[/tex]
    However, the correct answer is:
    [tex]4a^2(\frac{\pi}{2}-1)[/tex]

    OK, i think i understand. The surface area that i found was only one half of the circle for the top section of the "apple skin". But i need to find the sum of the top and bottom, so it's the answer that i got, multiplied by 4. Correct?

    By the way, is this the simplest/best method to get the answer? I seriously doubt that i will have enough time to do it in the exams... it just took me a whole day to reach an answer.
     
    Last edited: Jan 3, 2012
  15. Jan 3, 2012 #14
    Yes.

    The "coolness" of the final answer would suggest that there might be a simple geometrical argument, but I can't think of one. This method seems like a lot of work (because you've been working on the problem for a long time), but it really isn't. Noting that the integral transfers fairly well to polar coordinates makes the actual integration quite reasonable.
     
  16. Jan 3, 2012 #15

    ehild

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    Well done!

    It was not that complicated, was it? And it would have needed shorter time if you'd drawn a picture at the beginning, (or made holes in apples :wink:)

    ehild
     
  17. Jan 6, 2012 #16

    sharks

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    Thanks a lot, ehild and Screwdriver. Your advice has been crucial in helping me to understand this problem.

    Regards!
     
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