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1.
Find the area of the shaded region.
The shaded region can be divided into
rectangles along the y-axis. The area is the
integral of the difference between the right and
left functions.
7
2
π΄ππππππ = β« [(3π¦ β π¦ 2 ) β (π¦ 2 β 4π¦)]ππ¦
0
7
2
π΄ππππππ = β« (β2π¦ 2 + 7π¦)ππ¦
0
π΄ππππππ
π΄ππππππ
π΄ππππππ
π΄ππππππ
7
2π¦ 3 7π¦ 2 2
= [β
+
]
3
2
0
7 3
7 2
2( )
7( )
2(0)3 7(0)2
= β 2 + 2 β (β
+
)
3
2
3
2
2(343
7(49
)
8 )
=β
+ 4
3
2
343
=
π ππ’πππ π’πππ‘π
24
2.
Derive the volume of a sphere by revolving a semi-circle defined by π¦ = βπ 2 β π₯ 2 about the
x-axis where r is the (fixed) radius of the semi-circle.
Divided the sphere into cylindrical elements along the y-axis with radius π and altitude ππ¦. The
volume of this cylindrical element is ππ = ππ₯ 2 ππ¦. The sum of the cylindrical elements from 0 to r
π
is a hemisphere, so twice this value gives the sphere volume, or ππ πβπππ = 2(β«0 ππ₯ 2 ππ¦). Now,
we get the value of x in terms of y and substitute this to the equation for the volume.
π¦ = βπ 2 β π₯ 2
π₯2 = π2 β π¦2
π
ππ πβπππ = 2 (β« π(π 2 β π¦ 2 )ππ¦)
0
π
ππ πβπππ = 2π (β« (π 2 β π¦ 2 )ππ¦)
0
ππ πβπππ
π¦3 π
= 2π [π π¦ β ]
3 0
2
ππ πβπππ = 2π [((π 2 )(π) β
ππ πβπππ = 2π [π 3 β
2π 3
ππ πβπππ = 2π [
]
3
4ππ 3
ππ πβπππ =
3
π3
]
3
(π)3
(0)3
) β ((π 2 )(0) β
)]
3
3
3.
Find the volume of the cap of the sphere of radius r. The height of the cap is h.
The volume of the cap of the sphere is the sum of the cylindrical elements (with radius x) from the
base of the cap at π¦ = π β β to the top at οΏ½...