Previous exam questions on area between functions and volumes of solids.

1.Let f(x) = cos(x2) and g(x) = ex, for –1.5 ≤ x ≤ 0.5.
Find the area of the region enclosed by the graphs of f and g. (Total 6 marks)

2.Let f(x) = Aekx + 3. Part of the graph of f is shown below.

The y-intercept is at (0, 13).

(a)Show that A =10.
(2)

(b)Given that f(15) = 3.49 (correct to 3 significant figures), find the value of k. (3)

(c)(i)Using your value of k, find f′(x).
(ii)Hence, explain why f is a decreasing function.
(iii)Write down the equation of the horizontal asymptote of the graph f. (5)

Let g(x) = –x2 + 12x – 24.
(d)Find the area enclosed by the graphs of f and g.
(6)
(Total 16 marks)

3.The following diagram shows the graphs of f (x) = ln (3x – 2) + 1 and g (x) = – 4 cos (0.5x) + 2, for 1 £ x £ 10.

(a)Let A be the area of the region enclosed by the curves of f and g. (i)Find an expression for A.
(ii)Calculate the value of A.
(6)

(b)(i)Find f ′ (x).
(ii)Find g′ (x).
(4)

(c)There are two values of x for which the gradient of f is equal to the gradient of g. Find both these values of x. (4)
(Total 14 marks)

4.The graph of f(x) = , for –2 ≤ x ≤ 2, is shown below.

The region enclosed by the curve of f and the x-axis is rotated 360° about the x-axis. Find the volume of the solid formed.
(Total 6 marks)

5.The graph of y = between x = 0 and x = a is rotated 360° about the x-axis. The volume of the solid formed is 32π. Find the value of a. (Total 7 marks)

...Surface Area to Volume Ratio and the Relation to the Rate of Diffusion
Aim and Background
This is an experiment to examine how the Surface Area / Volume Ratio affects the rate of diffusion and how this relates to the size and shape of living organisms.
The surface area 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 surface area 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 Surface Area/Volume Ratio. The variable being tested in this experiment is the rate of diffusion in relation to the size of the gelatin cube. Another...

...Investigating Ratios of Areas and Volumes
In this portfolio, I will be investigating the ratios of the areas and volumes formed from a curve in the form y = xn between two arbitrary parameters x = a and x = b, such that a < b. This will be done by using integration to find the area under the curve or volume of revolution about an axis. The two areas that will be compared will be labeled ‘A’ and ‘B’ (see figure A). In order to prove or disprove my conjectures, several different values for n will be used, including irrational, real numbers (π, √2). In addition, the values for a and b will be altered to different values to prove or disprove my conjectures. In order to aid in the calculation, a TI-84 Plus calculator will be used, and Microsoft Excel and WolframAlpha (http://www.wolframalpha.com/) will be used to create and display graphs.
Figure A
1. In the first problem, region B is the area under the curve y = x2 and is bounded by x = 0, x = 1, and the x-axis. Region A is the region bounded by the curve, y = 0, y = 1, and the y-axis. In order to find the ratio of the two areas, I first had to calculate the areas of both regions, which is seen below. For region A, I integrated in relation to y, while for region B, I integrated in relation to x. Therefore, the two formulas that I used were y = x2 and x = √y, or x = y1/2....

...Mathematics
Volume of Solids
Formulae for Volume of Solids
Cube | Cuboid | Triangular Prism | Cylinder | Cone | Pyramid | Sphere | AnyPrism |
s3 | lwh | ½bhl | Πr2h | 1/3πr2h | 1/3Ah | 4/3πr3 | Ah |
A = area of the base of the figure
s = length of a side of the figure
l = length of the figure
w = width of the figure
h = height of the figure
π = 22/7 or 3.14
1. Compute the volume of a cube with side 7cm.Volume of cube: s3
s = 7cm
s3 = (7cm x 7cm x 7cm) = 343cm3
2. Compute the volume of a cuboid (also known as a rectangular prism) with the dimensions 4cm by 13cm by 9cm.
Volume of a cuboid: lwh
l = 4cm
w = 13cm
h = 9cm
lwh = (4cm x 13cm x 9cm) = 468cm3
3. Compute the volume of a triangular prism with a base length of 60cm, a base width of 8cm, and a height of 10cm.
Volume of a triangular prism: ½bhl
½b = (8cm x 1/2) = 4cm
h = 10cm
l = 60cm
½bhl = (4cm x 10cm x 60cm) = 2400cm3
4. Compute the volume of a cylinder which is 2m tall and has a radius 75cm. Convert this litres.
Volume of a cylinder: πr2h
π = 3.1415
r2 = (75cm)2 = 375 cm2
h = 2m = 200cm
πr2h = 235612. 5 cm3
cm L = 1cm 0.001 L
235612.5 cm3/ 1000 = 235.6125 L
5. Compute the volume of a cone with a radius of 200cm and a height of 0.75m.
Volume of a cone: 1/3πr2h
1/3π = 1.047
r2 =...

...Surface area / Volume ratio Experiment
Introduction:
The surface area 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 surface area 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. In the alkali...

...Introduction
Cells have to interact with their environment, chemicals and water and in order to do so they must be able to move across the cell membrane and the cell. The movements within a cell are called Diffusion. When molecules move across a cell membrane it is known as Osmosis. Diffusion is the process by which molecules of a substance move from areas of higher concentration of that substance to areas of lower concentration. Diffusion can be the transfer of anything anywhere. However, that is not true for osmosis. Osmosis is diffusion, but a specific type of diffusion. Osmosis is only the diffusion of water molecules through a selectively permeable membrane and a concentration gradient. Only certain molecules can cross the membrane to either go in our out of the cell.
If outside the cell there is a lower concentration of molecules than inside the cell, it is called hypotonic and water will move from outside the cell into the cell. If the solution outside the cell has a larger concentration of dissolved materials the solution is hypertonic, so the water would move from the cell into the solution.
Aim
The aim of this experiment was to verify the concept of Osmosis and Diffusion with a semi-permeable membrane (dialysis tubing), it will be exposed to different environments and concentration gradients.
Hypothesis
I thought that that dialysis tubing would end up weighing more as there would be less water and more molecules...

...Mathematics Class X (Term–II)
13
SURFACE AREAS AND VOLUMES
A. SUMMATIVE ASSESSMENT
(c) Length of diagonal =
TH
G
(a) Lateral surface area = 4l2 (b) Total surface area = 6l2 (c) Length of diagonal = 3 l 3. Cylinder : For a cylinder of radius r and height h, we have : (a) Area of curved surface = 2πrh
BR
O
ER
2. Cube : For a cube of edge l, we have :
O
YA L
TEXTBOOK’S EXERCISE 13.1 Unless stated otherwise, take =
22 . 7
Q.1. 2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid. [2011 (T-II)]
1
S
l 2 b2 h2
5. Sphere : For a sphere of radius r, we have : Surface area = 4πr2 6. Hemisphere (solid) : For a hemisphere of radius r we have : (a) Curved surface area = 2πr2 (b) Total surface area = 3πr2
PR
(a) Lateral surface area = 2h(l + b) (b) Total surface area = 2(lb + bh + lh)
(d) Total surface area of hollow cylinder = 2πh(R + r) + 2π(R2 – r2) 4. Cone : For a cone of height h, radius r and slant height l, we have : (a) Curved surface area = πrl = r h 2 r 2 (b) Total surface area = πr2 + πrl = πr (r + l)
Sol. Let the side of cube = y cm
Volume of cube = 64 cm3 Then, volume of cube = side3 = y3 As per condition y3 = 64 y3...

...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 surface area 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...