# Surface area - Double integrals

• etf
In summary: Spherical coordinates are nice because they allow for easy integration.In summary, the sphere has a surface area of 180.0261 between z = -2√3 and z = 2.
etf
Hi!
Calculate surface area of sphere $$x^{2}+y^{2}+z^{2}=16$$ between $$z=2$$ and $$z=-2\sqrt{3}$$.
Here are 3D graphs of our surfaces:

Surface area of interest is P3. It would be P-(P1+P2), where P is surface area of whole sphere. Is it correct?
Here is how I calculated P1:

Similarly for P2:

We calculate P using formula 4pi*r*r.
Is it correct?

P1=8*pi*ln2, P2=2*pi*ln(16/9), P=4*pi*16
P=4*pi*16-(8*pi*ln2+2*pi*ln(16/9))=201.0619-21.0358=180.0261

etf said:
P1=8*pi*ln2, P2=2*pi*ln(16/9), P=4*pi*16
P=4*pi*16-(8*pi*ln2+2*pi*ln(16/9))=201.0619-21.0358=180.0261

http://en.wikipedia.org/wiki/Spherical_segment

It seems there is a disagreement somewhere in your calculations between the formula and your result.

etf, have you given any consideration to working the problem in spherical coordinates? You can calculate it directly in about two steps.

I'm not familiar with spherical coordinates :(

Doing multiple integrals and not being familiar with integral transforms is not a good thing. Polar, spherical and cylindrical co-ordinates are important to be able to integrate properly, and without them, you won't get very far.

An interesting way to do this problem is to describe each face of the surface and then perform a surface integral over each one:

$$\iint_S \space dS = \iint_{S_1} \space dS_1 + \iint_{S_2} \space dS_2 + \iint_{S_3} \space dS_3$$

So you have a flat top surface ##S_1## described by the plane ##z = 2## and bounded by the region ##D_1## given by ##x^2 + y^2 = 12##. You also have a flat bottom surface ##S_2## describe by the plane ##z = -2 \sqrt{3}## and bounded by the region ##D_2##. Finally, you have the spherical middle section bounded by the planes.

Zondrina said:
Doing multiple integrals and not being familiar with integral transforms is not a good thing. Polar, spherical and cylindrical co-ordinates are important to be able to integrate properly, and without them, you won't get very far.

An interesting way to do this problem is to describe each face of the surface and then perform a surface integral over each one:

$$\iint_S \space dS = \iint_{S_1} \space dS_1 + \iint_{S_2} \space dS_2 + \iint_{S_3} \space dS_3$$

So you have a flat top surface ##S_1## described by the plane ##z = 2## and bounded by the region ##D_1## given by ##x^2 + y^2 = 12##. You also have a flat bottom surface ##S_2## describe by the plane ##z = -2 \sqrt{3}## and bounded by the region ##D_2##. Finally, you have the spherical middle section bounded by the planes.
I think the OP was supposed to calculate only the area of the spherical surface lying between the planes z = -2√3 and z = 2. I don't think the areas of S1 or S2 were ever intended to be included.

SteamKing said:
I think the OP was supposed to calculate only the area of the spherical surface lying between the planes z = -2√3 and z = 2. I don't think the areas of S1 or S2 were ever intended to be included.

Sorry, I saw the problem a little differently when I first read it this morning. The sphere with its ends cut off seems to be more appropriate, so the ##dS_3## integral would be the only thing required. I believe the conventional method of evaluation then proceeds as usual:

$$\iint_S \space dS = \iint_D \sqrt{z_x^2 + z_y^2 + 1} \space dA$$

The two planes and sphere give two circles in the x-y plane from which the limits can be deduced.

Zondrina said:
Sorry, I saw the problem a little differently when I first read it this morning. The sphere with its ends cut off seems to be more appropriate, so the ##dS_3## integral would be the only thing required. I believe the conventional method of evaluation then proceeds as usual:

$$\iint_S \space dS = \iint_D \sqrt{z_x^2 + z_y^2 + 1} \space dA$$

The two planes and sphere give two circles in the x-y plane from which the limits can be deduced.

That is certainly not the "natural" method to use. The surface is not symmetric vertically so you would have to work the top and bottom portions separately. And rectangular coordinates also are not the natural choice.

## 1. What is the formula for finding surface area using double integrals?

The formula for finding surface area using double integrals is ∭∭f(x,y)√(1+(∂f/∂x)^2+(∂f/∂y)^2)dA, where f(x,y) represents the function describing the surface and dA represents the infinitesimal area element.

## 2. How is a double integral used to find surface area?

A double integral is used to find surface area by integrating the function representing the surface over a specified region in the x-y plane. This integrates the function in both the x and y directions, resulting in a surface area.

## 3. What is the difference between a single integral and a double integral for finding surface area?

A single integral is used to find the area under a curve in one dimension, while a double integral is used to find the surface area of a 3-dimensional object in two dimensions. A single integral integrates over a line, while a double integral integrates over a region in the x-y plane.

## 4. Can double integrals be used to find surface area for any 3-dimensional object?

Yes, double integrals can be used to find surface area for any 3-dimensional object that can be described by a function in the x-y plane. This includes objects such as spheres, cones, and cylinders.

## 5. How is the accuracy of surface area calculated using double integrals?

The accuracy of surface area calculated using double integrals depends on the number of iterations used in the integration process. The more iterations, the more accurate the result will be. Additionally, using more precise mathematical methods, such as numerical integration, can also improve accuracy.

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