# Calculating Surface Area & Volume of 3D Function

• bomba923
In summary: I tried to numerically integrate the equation but got an elliptical region with no ending. I then tried z=b but got the same result. Is there a way to restrict the integration to a specific interval instead?)Area/Volume QuestionGiven the function on a 3D coordinate system:y = - z\sin \left( {xz} \right) where \left| {x - \frac{\pi }{2}} \right| \leqslant \frac{\pi }{z},\;z \ne 0and
bomba923
Area/Volume Question

Given the function on a 3D coordinate system:
$$y = - z\sin \left( {xz} \right)$$ where $$\left| {x - \frac{\pi } {2}} \right| \leqslant \frac{\pi }{z},\;z \ne 0$$

How do find the surface area of the figure integrated from z=a to z=b (where 'a' and 'b' are constants) ?

How would you also find the volume of this figure, from z=a to z=b (well, bounded by the top and by the trough, that is ) ?

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That interval on "z" induces an interval on "x" (by the modulus inequality).So you have your limits to the Riemann version of the surface integral.
U got the surface equation in explicit form

$$y(z,x)=-z\sin xz$$

U must know the formula giving an area for a surface in $\mathbb{R}^{3}$ whose equation is given explicitely.

I don't promiss the integration will be easy.

Daniel.

Guys,

I'm not sure, but it looks like this:

$$\int_{z_a}^{z_b}\int_{\frac{\pi}{2}-\frac{\pi}{z}}^{\frac{\pi}{2}+\frac{\pi}{z}} \sqrt{z^4Cos^2(xz)+(zxCos(xz)+Sin(xz))^2+1}dxdz$$

Edit: The surface area I mean.
Well, I guess I should explain to you how I arrived at this but I'll need to verify it before I do and perhaps others can provide additional input in the interim period . . .

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If it's really like that,then u can say good-bye from the integration.It's very elliptical...

Daniel.

saltydog said:
Guys,

I'm not sure, but it looks like this:

$$\int_{z_a}^{z_b}\int_{\frac{\pi}{2}-\frac{\pi}{z}}^{\frac{\pi}{2}+\frac{\pi}{z}} \sqrt{z^4Cos^2(xz)+(zxCos(xz)+Sin(xz))^2+1}dxdz$$

Edit: The surface area I mean.
Well, I guess I should explain to you how I arrived at this but I'll need to verify it before I do and perhaps others can provide additional input in the interim period . . .

Ok, I think I'm wrong: The function f(x,z) oscillates above and below the x-z plane. The formula I'm using above works only for a surface above the x-z plane. Requires further study . . .

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Ok, I think I'm wrong: The function f(x,z) oscillates above and below the x-z plane. The formula I'm using above works only for a surface above the x-z plane. Requires further study . . .

Why does that matter? The surface area of y = f(x, z) and of y = 10^10^100 + f(x, z) is the same, gives the same formula, and the latter does not go below the x-z plane.

saltydog said:
The function f(x,z) oscillates above and below the x-z plane

$$\left| {x - \frac{\pi }{2}} \right| \leqslant \frac{\pi }{z},\;z \ne 0$$ constricts $$y(x,z)$$ to a half-cycle, but I don't think it will oscillate. I mean, this given interval means half-cycle is either above the xz-plane or below the xz-plane (no "oscillation" between), depending on the sign of 'z' (did that to simplify to a half-cycle)--but not on both sides (unless you widen the interval, which might, i think, make it more difficult with oscillation)

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bomba923 said:
$$\left| {x - \frac{\pi }{2}} \right| \leqslant \frac{\pi }{z},\;z \ne 0$$ constricts $$y(x,z)$$ to a half-cycle, but I don't think it will oscillate. I mean, this given interval means half-cycle is either above the xz-plane or below the xz-plane (no "oscillation" between), depending on the sign of 'z' (did that to simplify to a half-cycle)--but not on both sides (unless you widen the interval, which might, i think, make it more difficult with oscillation)

Ok, thanks. I was looking at the whole plot without restricting it to the area between the lower and upper curve in z. Sooooo . . . maybe I do have it right. I'll spend some time on it and will ultimately just NIntegrate it in Mathematica and report the results. Can you report your results also?

I've included a plot of the boundary of the integration limits. I'll numerically integrate it from say z=1 to z=5 between the lower and upper curve and report the results.

#### Attachments

• surfaceint.JPG
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saltydog said:
$$\int_{z_a}^{z_b}\int_{\frac{\pi}{2}-\frac{\pi}{z}}^{\frac{\pi}{2}+\frac{\pi}{z}} \sqrt{z^4Cos^2(xz)+(zxCos(xz)+Sin(xz))^2+1}dxdz$$
I think followed your approach to solve it--find the dL (arc length) for the y-curve with respect to x (using dx), next multiply by dz to get dA, and then use a double integral to solve it. However, I'm not quire sure if my approach is correct ...(and thus the thread-!)

(My goal here for this problem is to find a general form (for the area And volume), given any bivariate function y(x,z) in 3D with x-interval [f(x,z), g(x,z)], integrated across from z=a to z=b. I started out here with a specific case :shy: )

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bomba923 said:
I think followed your approach to solve it--find the dL (arc length) for the y-curve with respect to x (using dx), next multiply by dz to get dA, and then use a double integral to solve it. However, I'm not sure if my approach is correct...(and thus the thread-!)

(My goal here for this problem is to find a general form, given any bivariate function y(x,z) in 3D with x-interval [f(x,z), g(x,z)], integrated across the z-axis. I started out here with a specific case :shy: )

Well, I'm just using the standard theorem for calculating surface area; it should be in your Calculus book. The one I'm looking at now is Leithold, p. 995. Anyway, I got a surface area between z=1 and z=5 of 59.135. I must tell you I'm not confident of it withoug further study of the matter however.

Neither am I :shy:
I got same answer (well, first I wrote it wrong, but then I fixed it and got 59.1)
Check out the 3D graph--it looks kind of neat

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Hmm---it seems the domain of my equation was wrong from the start! In addition, (as it is a periodic function), the whole problem can actually be simplified as
$$y\left( {x,z} \right) = z\cos \left( {zx} \right),\left| {zx} \right| \leqslant \frac{\pi }{2}$$
*And then one finds the surface area generated from z=a to z=b (across a much nicer interval)

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bomba923 said:
Hmm---it seems the domain of my equation was wrong from the start! In addition, (as it is a periodic function), the whole problem can actually be simplified as
$$y\left( {x,z} \right) = z\cos \left( {zx} \right),\left| {zx} \right| \leqslant \frac{\pi }{2}$$
*And then one finds the surface area generated from z=a to z=b (across a much nicer interval)

Ok, in that case I get:

$$\int_{z_a}^{z_b}\int_{-\frac{\pi}{2z}}^{\frac{\pi}{2z}} \sqrt{z^4Cos^2(xz)+(Cos(xz)-zxSin(xz))^2+1}dxdz$$

For the interval z=1 to z=5, I get 26.49.

## 1. How do you calculate the surface area of a 3D function?

The surface area of a 3D function can be calculated by taking the integral of the function over the desired range of values. This means breaking the function into small pieces and finding the area of each piece, then adding them together.

## 2. What is the formula for finding the volume of a 3D function?

The formula for finding the volume of a 3D function is similar to that of finding the surface area. It involves taking the integral of the function over the desired range of values and adding up the volumes of each small piece.

## 3. Can the surface area and volume of a 3D function be calculated without using calculus?

No, calculus is necessary for calculating the surface area and volume of a 3D function. The process involves finding the area or volume under a curve, which is a fundamental concept in calculus.

## 4. How does the shape of a 3D function affect its surface area and volume?

The shape of a 3D function directly affects its surface area and volume. A more complex or irregular shape will have a larger surface area and volume compared to a simpler or more uniform shape.

## 5. Are there any real-world applications for calculating the surface area and volume of 3D functions?

Yes, there are many real-world applications for calculating the surface area and volume of 3D functions. It is commonly used in engineering, architecture, and physics to design and analyze structures and objects in three-dimensional space.

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