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Volume under a surface

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  • #1
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Homework Statement


calculating volume under a surface, defined by implicit function f(x, y, z)=0 (in cartesian coordinates, strictly not in polar). Because the function that i need to integrate is quite complicated and there would be no obvious way to double check the result i first tried to calculate volume under sphere, but i get the wrong result.


Homework Equations


so let it be [tex]f(x, y, z)=x^2+y^2+z^2-1[/tex]


The Attempt at a Solution


[tex]V=8*\int dxdydz[/tex]
where dz is integrated from 0 to [tex]\sqrt{1-x^2-y^2}[/tex], dy from 0 to 1 and the same with dx. The multiplication factor 8 is added to get the whole volume.
I use Mathematica for solving the equation and get [tex]\frac{2}{3}(2\pi - i*(-4 + \log{16}))[/tex]

what am I doing wrong?
 

Answers and Replies

  • #2
33,167
4,851

Homework Statement


calculating volume under a surface, defined by implicit function f(x, y, z)=0 (in cartesian coordinates, strictly not in polar). Because the function that i need to integrate is quite complicated and there would be no obvious way to double check the result i first tried to calculate volume under sphere, but i get the wrong result.


Homework Equations


so let it be [tex]f(x, y, z)=x^2+y^2+z^2-1[/tex]


The Attempt at a Solution


[tex]V=8*\int dxdydz[/tex]
where dz is integrated from 0 to [tex]\sqrt{1-x^2-y^2}[/tex], dy from 0 to 1 and the same with dx. The multiplication factor 8 is added to get the whole volume.
I use Mathematica for solving the equation and get [tex]\frac{2}{3}(2\pi - i*(-4 + \log{16}))[/tex]

what am I doing wrong?
This should work for you.
[tex]8 \int_{x = 0}^1 \int_{y = 0}^{\sqrt{1 - x^2}} \int_{z = 0}^{\sqrt{1 - x^2 - y^2}} dz~dy~dx[/tex]

The first (inner) integration calculates the volume of a stack of small cubes, from the x-y plane up to the surface. The middle integration makes a wall of those stacks, going from y = 0 to the circle in the x-y plane. The final (outer) integration integrates those walls to give the volume of 1/8 of the sphere.
 
  • #3
Char. Limit
Gold Member
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Wouldn't it be easier, since you're integrating a sphere, to use spherical coordinates?
 
  • #4
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Thanks, Mark.

@Char. Limit
Yes, it would be easier in this case, but not in general (sphere served only as an example)
 
  • #5
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I'd have one more question regarding this sphere example - how could I rigorously calculate the surface area?
If I got the formula correctly, this could be done with:
http://en.wikipedia.org/wiki/Area - general formula (bottom of the page)

again in the case of sphere: [tex] f(x, y)=R^{2}-x^{2}-y^{2} [/tex]
and the boarders are determined by [tex] x^{2}+y^{2}=R^{2} [/tex]

but I still don't get the predicted result - [tex]4\pi R^2[/tex]

I could really use some help again:)
 
  • #6
33,167
4,851
I'd have one more question regarding this sphere example - how could I rigorously calculate the surface area?
If I got the formula correctly, this could be done with:
http://en.wikipedia.org/wiki/Area - general formula (bottom of the page)

again in the case of sphere: [tex] f(x, y)=R^{2}-x^{2}-y^{2} [/tex]
The graph of this function is not a sphere - it's a paraboloid. For a sphere of radius R, centered at (0, 0, 0), the equation is x2 + y2 + z2 = R2.

If you solve for z, you get
[tex]z = \pm \sqrt{R^2 - x^2 - y^2}[/tex]

If you define f(x, y) to be the positive square root of the above, you get a function whose graph is the upper hemisphere.
and the boarders are determined by [tex] x^{2}+y^{2}=R^{2} [/tex]
I think you mean borders:wink:
but I still don't get the predicted result - [tex]4\pi R^2[/tex]

I could really use some help again:)
 
  • #7
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back again (thanks Mark44 for fas reply), but I still have questions.
Let's say that I have a surface defined by [tex]x^2+y^2=f(z)[/tex] or rewritten [tex]x=\pm\sqrt{f(z)-y^2}=f(z, y)[/tex].
The formula for surface area is in this case: [tex]P=\int\int\sqrt{ (\frac{\partial f(y, z)}{\partial y})^2+(\frac{\partial f(y, z)}{\partial z})^2 +1 } dy dz[/tex],
where we first integrate [tex]dy[/tex] from [tex]y=\pm\sqrt{f(z)}[/tex].
The borders (:)) for [tex]dz[/tex] are then defined by the zeros of function [tex]f(z)[/tex].

is this OK?
 
  • #8
33,167
4,851
back again (thanks Mark44 for fas reply), but I still have questions.
Let's say that I have a surface defined by [tex]x^2+y^2=f(z)[/tex] or rewritten [tex]x=\pm\sqrt{f(z)-y^2}=f(z, y)[/tex].
I'm not sure what you mean above. The expression on the left is a function of x and y, but the one on the right is a function of z alone.

In your integral below you have f(y, z), so is f a function of one variable or two?
The formula for surface area is in this case: [tex]P=\int\int\sqrt{ (\frac{\partial f(y, z)}{\partial y})^2+(\frac{\partial f(y, z)}{\partial z})^2 +1 } dy dz[/tex],
where we first integrate [tex]dy[/tex] from [tex]y=\pm\sqrt{f(z)}[/tex].
The borders (:)) for [tex]dz[/tex] are then defined by the zeros of function [tex]f(z)[/tex].

is this OK?
 
  • #9
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Sorry, when I read it again I saw that it is unclear:
f(z) is any function, that depends on z - for example: [tex]x^2+y^2=(z-1)^2[/tex] - this is some given function.
In the case of f(y, z) I just wanted to stress the dependance and f(y, z) serves only as an abbreviation.
 
  • #10
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4,851
Instead of complicating things by having f(z) on one side and a different function of x and y on the other, can you just solve for z to get z in terms of x and y?

In your example, x2 + y2 = (z - 1)2, so

[tex]z - 1 = \pm \sqrt{x^2 + y^2}[/tex]
[tex]\Rightarrow z = 1 \pm \sqrt{x^2 + y^2}[/tex]

Here z is NOT a function, but the upper half of the graph is a function (of x and y), and the lower half is a function as well.

The upper half of the graph uses the positive square root; the lower half uses the negative square root.
 
  • #11
33,167
4,851
Looking at your example and previous posts about the iterated integral some more, you could have x as a function of y and z. Using your example, x = (z - 1)2 - y2. Solving for x gives
[tex]x = \pm \sqrt{(z - 1)^2 - y^2}[/tex]

Here, x is in terms of y and z, but is not a function. If we restrict x to be nonnegative, then we get a function:
[tex]x = + \sqrt{(z - 1)^2 - y^2} = f(y, z)[/tex]

For x <= 0, we get another function
[tex]x = - \sqrt{(z - 1)^2 - y^2} = -f(y, z)[/tex]
 

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