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Surface Area of a Sphere

  1. Nov 1, 2006 #1
    I've been looking at this method here:


    I was wondering at the last step before the "Note on multi-valuedness" if you wanted to obtain the general formula for the surface area of a sphere [tex]4 \times \pi \times r^2[/tex] with a radius of well r what limits would you use for each of the integrations? Well i say each of the integrations it looks like only one there (only one integration sign) but with the dx and dy after it does that mean it can be split up into two integrations? If so I don't understand where the [tex] \pi [/tex] comes from?

    I mean i'm guessing the f(x,y) and z cancel leaving just the r is that right?

    Sorry if this is a silly question!
  2. jcsd
  3. Nov 1, 2006 #2


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    Yes, it's a double integral over the region x^2+y^2<=r^2 in the xy plane. You can figure out what the bounds are. If all you want is the area, just take f=1 (or f=2 to account for the upper and lower hemispheres). The pi will come from doing the integral, which will involve a trig substitution.
  4. Nov 1, 2006 #3
    Oh i sort of get it, not just randomly, it's because of your reply (thanks :D) just drew a quick sketch i understand why

    x = rcos(a) and y = rsin(a)

    unless those aren't right in which case i don't understand it :P

    I tried putting the first integral limits as r and 0 and the second integral limits as r and [tex](r^2 - y^2)^\frac{1}{2}[/tex] with the second one being dx and the first being dy and then i tried substituting y = rsin(a) but yeah hasn't quite worked out yet. But is that sort of close to the right track?
    Last edited: Nov 1, 2006
  5. Nov 1, 2006 #4


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    Well, either use cartesion (x,y) or polar coordinates, not both. If you're going to use cartesian coordinates, then you're close with the limits you have there, but not quite right. y ranging between 0 and r will only cover half the circle, and x ranging from [itex]\sqrt{r^2-y^2}[/itex] to r will cover a region outside the circle (specifically, right now you're integrating over the half of the region outside the circle and inside a square that it's inscribed in). On the other hand, if you want to do polar coordinates, you'll have to use the jacobian for the transformation, as they mention in that article.
  6. Nov 2, 2006 #5
    Sorry to be such a bother i just can't figure out what the limits are here's what i did:

    [tex]\int\int r dxdy[/tex]

    [tex]\int r dx[/tex] upper limit = [tex]\sqrt{r^2-y^2}[/tex] lower limit = 0

    [tex]\int \sqrt{r^2-y^2} \times r dy[/tex] upper limit = r lower limit = 0

    Which comes out with an answer of [itex]\frac{1}{4} \times \pi \times r^3[/itex] i think. I know if these were the right limits that this would only cover part of the sphere so i'd have to multiply by some number (16? shouldn't it be 8?) to get it over the whole sphere but the [itex]r^3[/itex] is throwing me. I mean if i only want the surface area of this part of the sphere are these limits better or am i still outside the circle?

    Thanks for your continued support :)
    Last edited: Nov 2, 2006
  7. Nov 2, 2006 #6


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    Projecting the sphere [itex]x^2+ y^2+ z
    ^2= r^2[/itex] into the xy-plane, z= 0, gives the circle [itex]x^2+ y^2= r^2[/itex]. The point of the limits of integration is to "cover" that disk. If you want to use x as the outer variable of integration then x will have to vary from -r to r. for each x, then, since [itex]x^2+ y^2= r^2[/itex], [itex]y= \pm\sqrt{r^2- x^2}[/itex]. The limits you give will only cover the first quadrant of the circle (because of symmetry you can multiply by 4.)
    The integral will be
    [tex]\int_{x=-r}^r\int_{y= -\sqrt{r^2- x^2}}^{\sqrt{r^2- x^2}} f(x,y)dydx[/tex]

    Of course, you have to have the correct "differential of surface area". One way to do that is this: Think of [itex]F(x,y,z)= x^2+ y^2+ z^2= r^2[/itex] as a level surface for the function F(x,y,z). Then the gradient, [itex]\nabla F(x,y,z)= 2x\vec{i}+ 2y\vec{j}+ 2z\vec{k}[/itex] is a vector perpendicular to the sphere at each point and is a "vector differential of area". Since you want to integrate in the xy-plane, "normalize" by making the z component 1: divide the vector by 2z to get [itex]\frac{x}{z}\vec{i}+ \frac{y}{z}\vec{j}+ \vec{k}[/itex]. Now find the length of that:
    [tex]\sqrt{\frac{x^2}{z^2}+ \frac{y^2}{z^2}+ 1}= \sqrt{\frac{x^2+ y^2+ z^2}{z^2}}= \frac{r}{z}[/tex]
    The differential of surface area is [itex]\frac{r}{z}dydx[/itex]. Since [itex]z= \pm\sqrt{r^2- x^2- y^2}[/itex], use the positive z and multiply by 2:
    [tex]2\int_{x=-r}^r\int_{y=-\sqrt{r^2- x^2}}^{\sqrt{r^2-x^2}}\frac{r}{\sqrt{r^2- x^2- y^2}} dydx[/tex]
    You are going to need a couple of complicated trig substitutions to do that. (Added: Well, one complicated trig substitution and then everything reduces nicely!) A better way is to use parametric equations for the surface of the sphere.
    Use polar coordinates with [itex]\rho[/itex] set to the constant r:
    [itex]x= rcos(\theta)sin(\phi)[/itex], [itex]y= rsin(\theta)sin(\phi)[/itex], [itex]z= rcos(\phi)[/itex]. Then the position vector of a point on the surface of the sphere is
    [tex]rcos(\theta)sin(\phi)\vec{i}+rsin(\theta)sin(\phi)\vec{j}+z= rcos(\phi)\vec{k}[/tex]
    Differentiate that with respect to [itex]\theta[/itex]:
    and with respect to [itex]\phi[/tex]:
    [tex]rcos(\theta)cos(\phi)\vec{i}+ rsin(\theta)cos(\phi)\vec{j}- rsin(\phi)\vec{k}[/tex]

    The "fundamental vector product" of the surface is the cross product of those two vectors:
    and the length of that is [itex]r^2sin(\phi)[/itex]. The "differential of surface area" in those parameters is [itex]r^2sin(\phi)d\theta d\phi[/itex].

    To cover the entire surface, [itex]\theta[/itex] must vary from 0 to [itex]2\pi[/itex] and [itex]\phi[/itex] must vary from 0 to [itex]\pi[/itex]. The surface area is:
    [tex]r^2\int_{\theta= 0}^{2\pi}\int_{\phi=0}^\pi sin(\phi) d\phi d\theta[/tex]
    a much simpler integral.
  8. Nov 2, 2006 #7
    Thanks for your reply i just followed through your steps and although i'm not very good at visualising i can understand what you've shown me :D

    Again thanks StatusX and HallsofIvy :)
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