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......Integral Calculus - Mock Quiz 2 - 30 mins
Mehdi Garrousian
Jan 25, 2013
Mehdi Garrousian Calculus 141- Mock Quiz 2
...1 A wedge is cut out of a cylinder ofradius 5 by cutting at a 45◦ anglethrough the center of the base.Find the volume.(∠BAC = π/4 rad)
...2 (a) Use the half-angle formula to compute thedefinite integral.
∫ π
0 sin2 x dx(b) Use the method of washers to compute thevolume of the region bounded by y = sinx,y =
√x+ 1, x = 0 and x = π about the line
y = 4....3 Use cylindrical shells to find the volume of thesolid obtained by rotating the region boundedby y = xex
3
, x = 2 and the y-axis.Mehdi Garrousian Calculus 141- Mock Quiz 2
Solutions on the next page
Mehdi Garrousian Calculus 141- Mock Quiz 2
.. Volume of the wedge
The cross-sections are isoceles triangles where sidelengths are equal. Assuming B = (x, y),
AB = BC = y.
So, Area = y2/2. Also, x and y are related byx2 + y2 = 25, so y2 = 25− x2 and we get the formula
Area(x) =25− x2
2. The domain of x is [−5, 5].
The volume is
vol =
∫ 5
−5
25− x2
2dx =
∫ 5
0
(25− x2)dx.
(note: the integrand is even so it cancels with the 2 in denominator)
∫ 5
0
(25− x2)dx = 25x− x3
3
]50
= 125− (125/3) = (2/3)(125) = 250/3
Mehdi Garrousian Calculus 141- Mock Quiz 2
.. Volume of the wedge
The cross-sections are isoceles triangles where sidelengths are equal. Assuming B = (x, y),
AB = BC = y.
So, Area = y2/2. Also, x and y are related byx2 + y2 = 25, so y2 = 25− x2 and we get the formula
Area(x) =25− x2
2. The domain of x is [−5, 5].
The volume is
vol =
∫ 5
−5
25− x2
2dx =
∫ 5
0
(25− x2)dx.
(note: the integrand is even so it cancels with the 2 in denominator)
∫ 5
0
(25− x2)dx = 25x− x3
3
]50
= 125− (125/3) = (2/3)(125) = 250/3
Mehdi Garrousian Calculus 141- Mock Quiz 2
.. Volume of the wedge
The cross-sections are isoceles triangles where sidelengths are equal. Assuming B = (x, y),
AB = BC = y.
So, Area = y2/2. Also, x and y are related byx2 + y2 = 25, so y2 = 25− x2 and we get the formula
Area(x) =25− x2
2. The domain of x is [−5, 5].
The volume is
vol =
∫ 5
−5
25− x2
2dx =
∫ 5
0
(25− x2)dx.
(note: the integrand is even so it cancels with the 2 in denominator)
∫ 5
0
(25− x2)dx = 25x− x3
3
]50
= 125− (125/3) = (2/3)(125) = 250/3
Mehdi Garrousian Calculus 141- Mock Quiz 2
.. Washers
Part a) Recall cos 2θ = cos2 θ − sin2 θ = 1− 2 sin2 θ. As aconsequence, sin2 θ = 1
2 (1− cos 2θ).
∫ π
0
sin2 x dx =1
2
∫ π
0
(1− cos 2x)dx =1
2
(∫ π
0
dx− 1
2
∫ π
0
2 cos 2xdx)
=1
2
(x]π0− (1/2) sin 2x
]π0
)=
π
2
Part b) Let y1 =√x+ 1 (blue) and y2 = sinx
(red). The inner and outer radii are given by theblue and red curves, respectively.
R = 4− y2 = 4− sinx, r = 4−√x− 1 = 3−
√x
The volume is given by
vol =
∫ π
0
π(R2−r2)dx =
∫ π
0
π((4−sinx)2−(3−
√x)2)dx = . . . next page
Mehdi Garrousian Calculus 141- Mock Quiz 2
.. Washers
Part a) Recall cos 2θ = cos2 θ − sin2 θ = 1− 2 sin2 θ. As aconsequence, sin2 θ = 1
2 (1− cos 2θ).∫ π
0
sin2 x dx =1
2
∫ π
0
(1− cos 2x)dx =1
2
(∫ π
0
dx− 1
2
∫ π
0
2 cos 2xdx)
=1
2
(x]π0− (1/2) sin 2x
]π0
)=
π
2
Part b) Let y1 =√x+ 1 (blue) and y2 = sinx
(red). The inner and outer radii are given by theblue and red curves, respectively.
R = 4− y2 = 4− sinx, r = 4−√x− 1 = 3−
√x
The volume is given by
vol =
∫ π
0
π(R2−r2)dx =
∫ π
0
π((4−sinx)2−(3−
√x)2)dx = . . . next page
Mehdi Garrousian Calculus 141- Mock Quiz 2
.. Washers
Part a) Recall cos 2θ = cos2 θ − sin2 θ = 1− 2 sin2 θ. As aconsequence, sin2 θ = 1
2 (1− cos 2θ).∫ π
0
sin2 x dx =1
2
∫ π
0
(1− cos 2x)dx =1
2
(∫ π
0
dx− 1
2
∫ π
0
2 cos 2xdx)
=1
2
(x]π0− (1/2) sin 2x
]π0
)=
π
2
Part b) Let y1 =√x+ 1 (blue) and y2 = sinx
(red). The inner and outer radii are given by theblue and red curves, respectively.
R = 4− y2 = 4− sinx, r = 4−√x− 1 = 3−
√x
The volume is given by
vol =
∫ π
0
π(R2−r2)dx =
∫ π
0
π((4−sinx)2−(3−
√x)2)dx = . . . next page
Mehdi Garrousian Calculus 141- Mock Quiz 2
.. Washers
· · · =∫ π
0
π((4− sinx)2 − (3−
√x)2)dx
=π
∫ π
0
((16− 8 sin+ sin2 x)− (9− 6
√x+ x)
)dx
=π
(∫ π
0
(7− 8 sin+6√x− x)dx+
∫ π
0
sin2 xdx
)
=π
((7x+ 8 cosx+
6x3/2
3/2− x2
2)
]π0
+ π/2
)
=π(7π − 16 + 4
√π3 − π2
2+ π/2
)
Mehdi Garrousian Calculus 141- Mock Quiz 2
.. Shells
The general formula for volume by cylindrical shells is∫ b
a2πxf(x)dx.
In this case, we have∫ 2
0
2πx · xex3
dx = 2π(1
3)
∫ 2
0
3x2ex3
dx
The numbers in red are inserted to make the substitution easy. Letu = x3, then du = 3x2dx and the bounds change to 0 and 8. So, weget
(2π
3)
∫ 8
0
eudu = (2π
3)eu]80= (
2π
3)(e8 − 1)
Mehdi Garrousian Calculus 141- Mock Quiz 2