Surface area
Surface area is the measure of how much exposed area a solid object has, expressed in square units. Mathematical description of the surface area is considerably more involved than the definition of arc length of a curve. For polyhedra (objects with flat polygonal faces) the surface area is the sum of the areas of its faces. Smooth surfaces, such as a sphere, are assigned surface area using their representation as parametric surfaces. This definition of the surface area is based on methods of infinitesimal calculus and involves partial derivatives and double integration. General definition of surface area was sought by Henri Lebesgue and Hermann Minkowski at the turn of the twentieth century. Their work led to the development of geometric measure theory which studies various notions of surface area for irregular objects of any dimension. An important example is the Minkowski content of a surface.

Definition of surface area

While areas of many simple surfaces have been known since antiquity, a rigorous mathematical definition of area requires a lot of care. Surface area is an assignment of a positive real number to a certain class of surfaces that satisfies several natural requirements. The most fundamental property of the surface area is its additivity: the area of the whole is the sum of the areas of the parts. More rigorously, if a surface S is a union of finitely many pieces S1, …, Sr which do not overlap except at their boundaries then

Surface areas of flat polygonal shapes must agree with their geometrically defined area. Since surface area is a geometric notion, areas of congruent surfaces must be the same and area must depend only on the shape of the surface, but not on its position and orientation in space. This means that surface area is invariant under the group of Euclidean motions. These properties uniquely characterize surface area for a wide class of geometric surfaces called piecewise smooth. Such surfaces consist of finitely...

...Planning
Aim
To determine how the surfacearea of the tablets affects the rate of the reaction. To determine which form of tablets gives the biggest surfacearea resulting in the fastest reaction rate.
Investigation question:
What is the relationship between the total surfacearea of the tablets and the rate of the reaction?
Hypothesis:
The rate of reaction will be the fastest when the tablets crushed into powder as there is a bigger total surfacearea resulting in more effective collisions between particles.
Variables:
Independent variable: Different forms of tablets.
Dependant variable: Time the syringe took to stop moving as the tablets dissolve.
Fixed variables:
*External temperature
*volume of HCl
*Temperature –all 3 final runs were done on the same day so whether was not an issue and did not affect the results
*Use of catalyst – a catalyst was not used in any of the experiments
* Use the same person to observe the reaction because different people have different eyesight
Background information relating to the experiment
In this experiment we are looking at one effect that influences the rate of reaction , namely total surfacearea. The reaction rate (rate of reaction) or speed of reaction for a reactant or product in a particular reaction is defined as how fast or slow a reaction takes place....

...Surfacearea / Volume ratio Experiment
Introduction:
The surfacearea to volume ratio in living organisms is very important. Nutrients and oxygen need to diffuse through the cell membrane and into the cells. Most cells are no longer than 1mm in diameter because small cells enable nutrients and oxygen to diffuse into the cell quickly and allow waste to diffuse out of the cell quickly. If the cells were any bigger than this then it would take too long for the nutrients and oxygen to diffuse into the cell so the cell would probably not survive. Single celled organisms can survive as they have a large enough surfacearea to allow all the oxygen and nutrients they need to diffuse through. Larger multi celled organisms need specialist organs to respire such as lungs or gills.
Apparatus Needed For the Experiments:
1. Beakers
2. Gelatin blocks mixed containing universal indicator
3. 0.1M Hydrochloric acid
4. Stop Watch
5. Scalpel
6. Tile
7. Safety glasses
Method:
1. A block of gelatin which has been dyed with universal indicator should be cut into blocks of the following sizes (mm).
5 x 5 x 5
10 x 10 x 10
15 x 15 x 15
20 x 20 x 20
10 x 10 x 2
10 x 10 x 10 (Triangle)
10 x 15 x 5
20 x 5 x 5
The triangle is of the following dimensions.
The rest of the blocks are just plain cubes or rectangular blocks.
Universal indicator is a neutral indicator....

...INTERNATIONAL BACCALAUREATE
BIOLOGY INTERNAL ASSESMENT
DATA COLLECTION AND PROCESSING
CONCLUSION AND EVALUATION
RELATIONSHIP BETWEEN SURFACEAREA AND THE VOLUME OF A CELL, HOW IT AFFECTS THE RATE OF DIFFUSION
Nazirul Ibrahim
04.10.2011
DATA COLLECTION AND PROCESSING
We have determined the following variables:
* dV- the time for the color indicator phenolphthalein to reach the center of the cube (diffusion)
* iV- the size of the cube
* cV- the amount of Sodium Hydroxide the perfect dimensions of the cube, how the cube is fully submerged in the solution, the temperature of the cube and the constant surrounding temperature, the concentration of the NaOH solution
During this experiment of investigating the relationship between the surfacearea and the volume of a cell, the following was determined:
* The amount of time (sec) taken for the pink color to reach the center of the cube
* How ratio of the cube (the volume and surfacearea) affects the time of diffusion
* The ratio of the
With the results obtained accurately, calculations were made to find out the following:
* The mean of each group’s data
* The standard deviation of each group’s data
As each trial consist for 5 different sizes of the agar cubes (1cm, 1.5cm, 2cm, 2.5cm and 3cm) and that the trials are repeated 5 times, we expect that results will be as follows:...

...investigation is to prove the effect of increasing size on the efficiency of diffusion. Diffusion is the process that cells use to obtain oxygen, water and food. Also, how they lose waste substances, for example, urea and carbon dioxide. Basically, Diffusion is when particles move from an area of high concentration to an area of low concentration. The surfacearea to volume ratio of the cell is an important factor in diffusion. It is the effect of this factor that will be investigated in this practical.
Materials used in the practical consist of blocks of agar jelly containing indicator, a sharp knife, used to cut the agar jelly into the recommended sizes, diluted sulphuric acid (300mL) , ruler, to measure the sizes of the cubes, absorbant paper towel, three 250mL beakers, and a measuring cylinder.
To perform this experiment, the following procedures will need to be follow precisely:
Obtain three blocks of agar jelly. Carefully cut three cubes, as close as possible to 1 cm, 2 cm, and 3 cm in size. Try to be exact, as this will help in the calculations later. Record the exact dimensions of the cube, in a table (see back of booklet).
Calculate the volume, the surfacearea, and surfacearea/volume ratio for each cube. These will be known as the ‘outer’ cube dimensions.
Measure 100 mL of sulphuric acid into each of the three beakers.
CAUTION:...

...SurfaceArea to Volume Ratio and the Relation to the Rate of Diffusion
Aim and Background
This is an experiment to examine how the SurfaceArea / Volume Ratio affects the rate of diffusion and how this relates to the size and shape of living organisms.
The surfacearea to volume ratio in living organisms is very important. Nutrients and oxygen need to diffuse through the cell membrane and into the cells. Most cells are no longer than 1mm in diameter because small cells enable nutrients and oxygen to diffuse into the cell quickly and allow waste to diffuse out of the cell quickly. If the cells were any bigger than this then it would take too long for the nutrients and oxygen to diffuse into the cell so the cell would probably not survive.
Single celled organisms can survive as they have a large enough surfacearea to allow all the oxygen and nutrients they need to diffuse through. Larger multi-celled organisms need organs to respire such as lungs or gills.
Method
The reason I chose to do this particular experiment is because I found it very interesting and also because the aim, method, results- basically the whole experiment would be easily understood by the average person who knew nothing about SurfaceArea/Volume Ratio. The variable being tested in this experiment is the rate of diffusion in relation to the size of the gelatin...

...AREA
(i) The area of a rhombus is equal to the area of a triangle whose base and the corresponding altitude are 24.8 cm and 16.5 cm respectively. If one of the diagonal of the rhombus is 22 cm, find the length of the other diagonal.
(ii) The floor of a rectangular hall has a perimeter 250m. If the cost of paining the four walls at the rate of Rs 10 per m2 is Rs 1500. Find the height of the hall.
(iii) A room is half as long again as it is broad. The cost of carpeting the room at Rs 3.25 per m2 is Rs 175.50 and the cost of papering the walls at Rs 1.40 per m2 is Rs 240.80. If 1 door and 2 windows occupy 8m2, find the dimensions of the room.
(iv) A river 2m deep and 45m wide is flowing at the rate of 3 km per hour. Find the volume of water that runs into the sea per minute.
(v) A closed cylinder has diameter 8cm and height 10cm. Find its total surfacearea and volume.
(vi) The volume of a metallic cylinder pipe is 748cm3 . Its length is 14 cm and external diameter 18cm. Find its thickness.
(vii) A cylindrical bucket, 28cm in diameter 72cm high is full of water. The water is emptied into a rectangular tank, 66cm long and 28cm wide. Find the height of the water level in the tank.
(viii) A cylindrical tube, open at both ends, is made of metal. The internal diameter of the tube is 10.4cm and its length is 25cm. The thickness of the metal is 8mm...

...Kimberly M Dollar
000234333
EFT4: Math: Task 5: SurfaceArea of Cubes
Introducing SurfaceArea
For a fifth or sixth grade class to understand the concept surfacearea in relation to a cube they need to understand what a cube is first. They will learn that a cube is a special type of rectangular solid. The length, width, and height of a cube are exactly the same. After explaining what a cube is they will need to understand what it means to find the surfacearea. The surfacearea is not the same as finding the volume of a cube. The surfacearea is the area on the outside of a three-dimensional shape, like the cube. The surfacearea of a cube is six times the surfacearea of one side of the cube. There are six sides to one cube, after learning this about a cube the appropriate formula to find the surfacearea is:
Surfacearea of a cube=6s^2 (The “6” represents the number of sides; “s” represents one side of a cube; “^2” represents taking one side and timing it by itself; the end result gives the surfacearea of a cube).
Prerequisite Skills
The necessary prerequisite skills required to determine the surfacearea...

...have to be taken in and to also be removed. This is where the surfacearea to volume ratio comes into place; the reason why this ratio is so important is because the surfacearea of a cell essentially affects the rate of the transferring of useful substances (through diffusion and osmosis etc.) in and out of the organism. On the other hand, the total volume of the organism also affects the rate of the making of material inside the cell and the ability to hold all of the substances. Whilst organisms are slowly growing and developing day by day, the volume of the organism increases, but not to the same extent as the surfacearea; this is because the organisms’ surfacearea increases at a much slower rate than its volume. Through research and experiments, it is apparent that as the organism grows, its surfacearea to volume ratio slowly decreases, the table to the right also proves that this theory is true; meaning that it would become increasingly difficult for the organism to obtain the required nutrients and also expelling the wastes produced by metabolism. In the end, it becomes impossible for diffusion to occur efficiently, where the cells becomes too large so they would divide through the process of mitosis. For example, the alveoli in our lungs have a relatively high large surfacearea to volume ratio,...