Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 6.2 VOLUMES ¤ 275
51. x 2 + y 2 = r 2 ⇔ x 2 = r 2 − y 2
V = π
= π
r
r−h
r 2 − y 2 dy = π
r 3 − r3
3
r 2 y − y3
3
−
r 2 (r − h) −
= π 2
3 r3 − 1 3 (r − h) 3r 2 − (r − h) 2
r
r−h
(r − h)3
3
= 1 3 π 2r 3 − (r − h) 3r 2 − r 2 − 2rh + h 2
= 1 3 π 2r 3 − (r − h) 2r 2 +2rh − h 2
= 1 π 2r 3 − 2r 3 − 2r 2 h + rh 2 +2r 2 h +2rh 2 − h 3
3
= 1 π 3rh 2 − h 3
= 1 3 3 πh2 (3r − h), or, equivalently, πh 2 r − h
3
53. For a cross-section at height y, we see from similar triangles that α/2
b/2 = h − y
1
h ,soα = b − y h
Similarly, for cross-sections having 2b as their base and β replacing α, β =2b 1 − y
.So
h
V =
=
h
0
h
0
A(y) dy =
2b 2 1 − y h
h
b 1 − y
2b 1 − y
dy
0 h
h
2dy
h
=2b
2
1 − 2y
h + y2
dy
h 2
h
=2b 2 y − y2
h + y3
=2b 2 h − h + 1
3h h
2 3
0
0
= 2 3 b2 h [ = 1 Bh where B is the area of the base, as with any pyramid.]
3
55. A cross-section at height z is a triangle similar to the base, so we’ll multiply the legs of the base triangle, 3 and 4, by a
proportionality factor of (5 − z)/5.Thus,thetriangleatheightz has area
A(z) = 1
5 − z 5 − z
2 · 3 · 4 =6 1 − z 2,so
5 5
5
V =
5
0
A(z) dz =6
5
0
1 − z 2
0
dz =6
5
= −30 1
u3 0
= −30
− 1 3 1 3 =10cm
3
1
u 2 (−5 du)
u =1− z/5,
du = − 1 dz 5
.
57. If l is a leg of the isosceles right triangle and 2y is the hypotenuse,
then l 2 + l 2 =(2y) 2 ⇒ 2l 2 =4y 2 ⇒ l 2 =2y 2 .
V = 2
A(x) dx =2 2
A(x) dx =2 2 1
(l)(l) dx =2 2
−2 0 0 2
2 (4 − 0 x2 ) dx
=2 2
0
= 9 2
1
4 (36 − 9x2 ) dx = 9 2
4x −
1
3 x3 2
0 = 9 2
8 −
8
3
=24
0 y2 dx