Page 216 - How To Solve Word Problems In Calculus

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The total volume

2

2
V = π y − y 1 2 dx
2
0
2

2 2
2 2
= π (8 − x ) − (x ) dx
0
2

4
4
2
= π [64 − 16x + x − x ] dx
0
2

2
= π [64 − 16x ] dx
0
3 2
16x
= π 64x −
3
0

128
= π 128 − − 0
3
256π
=
3
9. At first, this appears to be a simple problem: Why not
simply subtract the volume of the cylinder from the volume of the
sphere? The figure below shows why this is not a valid approach,
as part of the sphere is eliminated when the hole is drilled.
3
2

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