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sub topic 3 8 Flipbook PDF
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3.8 Application of Integration Area and volume of bounded region
x-axis π Area = β«π₯π π¦ πΧ¬β¬ π Volume = β« π¦π πΧ¬β¬2 ππ₯
y-axis π Area = β«π¦π π₯ πΧ¬β¬ π Volume = β« π₯π πΧ¬β¬2 ππ¦
Example : a) Find ythe area for the x-axis
b) Find the area for the y-axis y
3
π₯ = π¦2 + 2
π¦ = π₯2 1
x
1
2
x
0
A=
2 β«Χ¬β¬1 π₯ 2
ππ₯ = 8 3
π₯3 2 3 1
1 3
A= 7 3
= β = π’πππ‘π 2
3 β«Χ¬β¬0 (π¦ 2 +2)
ππ₯ =
π¦3 3
+ 2π¦
3 0
= 15 β 0 = 15 π’πππ‘π 2
Example : c) Find the area for the x-axis y
A=
π¦ = 4π₯ β π₯ 2
4 β«Χ¬β¬0 (4π₯
β
π₯ 2 ) ππ₯
=
4π₯ 2 2
3 4 = 2 4 2β β 2 0 3 0
x 4
=
32 β 3
0=
32 π’πππ‘ 2 3
β
4 π₯3 3 0
3 0 2β 3
d) Given an equation, π¦ = π₯ 2 + 6. Find the area under the graph bounded by the curve, y-axis. The line π¦ = 12 and π¦ = 16. Solution : Area bounded π¦ = 12 and π¦ = 16. π₯ = (π¦ β 6)1/2 β΄
16 β«Χ¬β¬12 (π¦
β 6)1/2 ππ¦ = =
=
(π¦β6)3/2
16
3 2
12 2 [(16 β 6)3/2 β(12 β 3 2 [(10)3/2 β(6)3/2 ] 3 2 16.926 3
= = 11.284
6)3/2 ]
Example : e) Find the volume of the solid generated by revolving the region between the x-axis and the parabola for the x-axis π¦ = 4π₯ β π₯ 2 through a complete revolution about the x-axis.
Solution : Find the x-intercepts of π¦ = 4π₯ β π₯ 2 When π¦ = 0 ;
0 = 4π₯ β π₯ 2 , π₯ = 0 ππ π₯ = 4
4
V = π β«Χ¬β¬0 (4π₯ β π₯ 2 )2 ππ₯ 4
= π ΰΆ±(16π₯ 2 β 8π₯ 3 +π₯ 4 ) ππ₯ 0
=π
16π₯ 3 3
β
8π₯ 4 4
+
4 π₯5 =π 5 0
16π₯ 3 3
β 2π₯ 4 +
4 π₯5 5 0
=π
=π
16 4 3
3
16 4 3
3
5 4 16 4 4 β2 4 + β( 5 3
3
4 5 16 0 + β 5 3
3
β2 4
4
1024 1024 =π β 512 + 3 5 πππ = π
πππππ ππ
5 4 β2 4 4+ ) 5
β2 0
4
0 5 + 5
4
0
Exercise : a) Find the area of the region bounded by the parabola π¦ = 2π₯ 2 β 13π₯, the x-axis and the lines x = 0 and x =5.
b) Find the area of the region bounded by the parabola x = β(π¦ β 4)2 , the y-axis and the lines y = 0 and y = 4.
c) Figure shows a region bounded by the curve π¦ = 4π₯ β π₯ 2 , and the line 2π₯ + π¦ = 8. Determine the volume generated when the region R is rotated through 3600 about the x-axis. π¦
8 2π₯ + π¦ = 8 π
π¦ = 4π₯ β π₯ 2
π₯ 0
2
4
d) Figure shows a region which is enclosed by the graph of π¦ = π₯ + 1 and y = (π₯ β 1)2 . Compute the volume of solid revolution formed when the shaded region is rotated 3600 about x-axis. π¦ y = (π₯ β 1)2 π¦ =π₯+1
0
2
π₯
e) Calculate the generated volume between the curve π¦ = 5π₯ β π₯ 2 , x βaxis x = 3 and x = 5 on figure below.
π¦ π¦ = 5π₯ β π₯ 2
6
0
π₯ 3
5
The endβ¦β¦. Good luck