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Finding surface area using double integrals

  1. Nov 23, 2011 #1
    1. The problem statement, all variables and given/known data

    The portion of the paraboloid 2z=x^2+y^2 that is inside the cylinder x^2+y^2=8

    3. The attempt at a solution
    my attempt was that i would turn this into polar coordinates and solve that integral but is it right? I came up with
  2. jcsd
  3. Nov 23, 2011 #2

    Simon Bridge

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  4. Nov 23, 2011 #3


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    That integral isn't correct. Show us how you calculated dS.
  5. Nov 23, 2011 #4
    woops my limits arent right on that, it should go from 0 to sqrt(8). is that what is wrong?
  6. Nov 23, 2011 #5


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    That's one thing.
  7. Nov 24, 2011 #6
    well i know that a cylinder makes the dr integral go from 0 to sqrt(8), and I know for dθ it goes all the way around so i know thats from 0 to 2pi, and for my integral i just did (x^2+y^2)/2 and then converted that to polar so it would be r^2/2 and then wrote out the complete equation ie
  8. Nov 24, 2011 #7


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    That would be right if you were asked to calculate the volume under the paraboloid and above the xy plane. But the title of your thread says you are asked to get the surface area of the paraboloid inside the cylinder. That is why I keep asking you for dS.
  9. Nov 24, 2011 #8


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    Any surface can be written as a vector equation:
    [tex]\vec{r}(u,v)= x(u,v)\vec{i}+ y(u,v)\vec{j}+ z(u,v)\vec{k}[/tex]
    where u and v are parameters.

    The vectors
    [tex]\vec{r}_u(u,v)= x_u\vec{i}+ y_u\vec{j}+ z_u\vec{k}[/tex]
    [tex]\vec{r}_v(u,v)= x_v\vec{i}+ y_v\vec{j}+ z_v\vec{k}[/tex]

    are tangent vectors. Their cross product gives the "differential of surface area":
    [tex]dS= \left|\vec{r}_u\times \vec{r}_v\right|du dv[/tex]
  10. Nov 24, 2011 #9
    wow im an idiot. so the equation i found was
    sa=double integral of the sqrt((fx)^2 + (fy)^2 +1)

    so this was my equation
    with the respective integrals of course. is that right?
    and If it is, i know i have to convert it to polar, which looks terribly difficult
    Last edited: Nov 24, 2011
  11. Nov 24, 2011 #10

    Simon Bridge

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    I gave you a link which shows you how to work out the integrand and do the conversion.
    You have to scroll down to the bit where it shows you the example related to your problem.
    I suspect you are still thinking in terms of integrating some function provided. Get that out of your head - the dS takes care of that for you. (Either that or you forgot to do the cross product...)

    The thinking:
    In general - when you have a flat area to calculate, you stamp the area with small squares and count them. dS is that small area. If the rea is flat then the sides of the small squares are dx and dy then the surface area of each is dS=dxdy. The Area of the whole surface is the sum of all the little areas like this:

    [tex]\int_{Y} \int_{X} dxdy[/tex] ... where X and Y represent the limits of integration.
    Notice how there is apprently no function to integrate? The function in question is g(x,y)=1 - because the surface is flat, it's slope in the x and y directions is 1.

    The area below a function in the x-y plane is a special case of this that simplifies to a single integral - so it is easier to teach.
    As a double integral, the area under f(x) is:

    [tex]\int_{a}^{b} \int_{y=0}^{y=f(x)} 1dydx[/tex]

    If the area is not flat, then you project the squares onto the surface - so they look like parallelograms - the sides will be longer depending on the slope of the surface in their directions (hint: partial derivatives) and dS will be in terms of dxdy but modified for the area of a parallelogram. Which is basically what you've just done.

    But you can also do this in other coordinate systems.

    In polar coords, a small area at position [itex](r,\theta)[/itex] from the z axis, in a plane perpendicular to the z axis, which covers a distance [itex]\Delta r[/itex] in the [itex]r[/itex] direction, and an angle [itex]\Delta \phi[/itex] in the [itex]\phi[/itex] would be roughly [itex]dS=(r\Delta \phi) \Delta r[/itex] - can you see why? In the limit that [itex]\Delta \phi[/itex] and [itex]\Delta r[/itex] are very small, then this area is exact with:

    [tex]dS = r dr d\phi[/tex]

    BUT: the surface is not flat!
    A parabaloid about z in cylindrical-polar coords would be:


    So the slope varies in the r direction - making the sides of dS longer in the r direction.
    A quick sketch of the situation will help.

    (BTW: You'll have noticed how useful LaTeX is by now - it is really worth learning.)
    Last edited: Nov 25, 2011
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