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1. PART 3
Diagram 2 shows a parabolic satellite disc which is symmetrical at the y-axis. Given that the
diameter of the discs is 8m and the depth is 1m.
Diagram 2
a) Since the parabolic satellite disc is symmetrical at the y-axis, the curve y=f(x) can be
written as y=ax2
+c. it can be seen that the curve y=f(x) cuts the y-axis at the point (0,4).
Substitute (0,4) into y=ax2
+c and you will get y=ax2
+4
Substitute the point (4,5) into y=ax2
+4
Therefore, 5=a(42
)+4
a= 1
16
Therefore, y= f(x) is now written as y=1 x2
+4
16
So, f(x)= x2
+4
16
y
8 m
1 m
0
(4,5)
4
0 4
y = f(x)
x

2. b) To find the approximate area under a curve, we can divide the region into several
vertical strips, then we add up the areas of all the strips. Using a scientific calculator or
any suitable computer software, estimate the area bounded by the curve y=f(x) at (a),
the x-axis, x=0 and x=4.
When x=0, f(0)= 02
+ 4
16
f (0) = 4
When x= 0.5, f(0.5) = (0.5)2
+ 4
16
f (0.5) = 4.0156
When x=1, f(1) = 12
+ 4
16
= 4.0625
When x= 1.5 f (x) = 1.52
+ 4
16
= 4.1406
When x= 2, f (x) = 22
+ 4
16
= 4.25
When x= 2.5 f (x) = 2.52
+ 4
16
= 4.3905

3. When x= 3.0 f (x) = 3.02
+ 4
16
= 4.5625
When x= 3.5 f (x) = 3.52
+ 4
16
= 4.7656
Area of the first strip = 4m x 0.5m
= 2m2
Area of the second strip = 4.0156m x 0.5m
= 2.0078m2
Area of the third strip = 4.0625m x 0.5m
= 2.0313m2
Area of the fourth strip = 4.1406m x 0.5m
= 2.0703m2
Area of the fifth strip = 4.25m x 0.5m
= 2.125m2
Area of the sixth strip = 4.3906m x 0.5m
= 2.1953m2
Area of the seventh strip = 4.5625m x 0.5m
= 2.2813m2
Area of the eighth strip = 4.7656m x 0.5m
= 2.3828m2
Total area = 2+2.0078+2.0313+2.0703+2.125+2.1953+2.2813+2.3828
= 17.0938 m2

4. b) ii)
Area of the first strip = 4.0156m x 0.5m
= 2.0078m2
Area of the second strip = 4.0625m x 0.5m
= 2.0313m2
Area of the third strip = 4.1406m x 0.5m
= 2.0703m2
Area of the fourth strip = 4.25m x 0.5m
= 2.125m2
Area of the fifth strip = 4.3906m x 0.5m
= 2.1953m2
Area of the sixth strip = 4.5625m x 0.5m
= 2.2813m2
Area of the seventh strip = 4.7656m x 0.5m
= 2.3828m2
Area of the eighth strip = 5m x 0.5m
= 2.5m2
Total area = 2.0078+2.0313+2.0703+2.125+2.1953+2.2813+2.3828+2.5
= 17.5938 m2

5. b ) iii)
Area of the first and second strips= 4.0156m x 1 m
=4.0156 m2
Area of the third and fourth strips = 4.1406m x 1m
= 4.1406m2
Area of the fifth and sixth strips = 4.3906 m x 1m
=4.3906m2
Area of the seventh and eighth strips = 4.7656m x 1m
= 4.7656m2
Total area = 4.0156+4.1406+4.3906+
= 17.3124 m2

6. c) i) Calculate the area under the curve using intergration.
Area= ∫ 𝑦 d𝑥
4
0
= ∫ ( 𝑥2
/16 + 4
4
0
) dx
= [ x3
/ (16x3) +4x)4
0
= [43
/48 + (4x4)] – 0
= 17
1
3
𝑚2
ii) Compare your answer in c(i) with the values obtained in (b). hence, discuss which
diagram gives the best approximate area.
The diagram 3 (iii) gives the best approximate area, which is 17.3124m2
iii) Explain how you can improve the value in c(ii).
We can improve the value in c(ii) by having more strips from x=0, to x=4.
d) Calculate the volume of the satellite disc.
y =
𝑥2
16
+ 4
x2
= 16(y-4)
x2
=16y-64
Volume= π ∫ 𝑥2
d𝑦
5
4
= π ∫ (16y − 64)dy
5
4
= π [16/2y2
- 64y]5
4
= 8 π cm3