MATH 152 - Recitation Quiz 2 - Spring 2015
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1.
Find the volume of the solid obtained by rotating the region bounded by
y=
√
x − 1, x = 2, x = 5, y = 0;
about the
x-axis.
x-axis, we get a disk. Thus, we will use
A = πr2 . Here, the radius is the distance from the x-axis to the top of the
2
√
√
r (x) = x − 1. Therefore, the area of the disk is A = π x − 1 = π (x − 1) .
If we revolve a small rectangle in the region described about the
the formula for area of a circle:
rectangle.
In other words
Therefore, the following gives us the volume:
ˆ5
1
π (x − 1) dx = π x2 − x
2
V =
2
2.
5
=π
2
15
15π
25
− 5 − (2 − 2) = π
−0 =
2
2
2
units
3
.
Set up, but do not evaluate, an integral for the volume of the solid obtained by rotating
the region bounded by
y = cos x, y = 0, x = 0, x = π/2;
about
y = −1.
y = −1, we get a washer. Thus, we will use the
A = πR2 − πr2 = π R2 − r2 . The longer radius is the distance from y = −1
to the top of the rectangle which has coordinate y = cos x. Therefore, R (x) = cos x − (−1) = cos x + 1. The
smaller radius is the distance from y = −1 to the bottom of the rectangle which is on the x-axis. Therefore,
h
i
h
i
2
2
2
r (x) = 1. Therefore, A = π (cos x − 1) − (1) = π (cos x − 1) − 1 . Therefore, the following gives us the
If we revolve a small rectangle in the region described about
formula for area of a washer:
volume:
ˆπ/2 h
i
2
V =
π (cos x − 1) − 1 dx.
0
3.
Set up, but do not evaluate, an integral for the volume of a solid
the base of
S
is a circular disk with radius
r.
S
obtained as follows:
Parallel cross-sections perpendicular to the
base are squares. HINT: The equation of the top half of a circle of radius
r
centered at
the origin is
p
r 2 − x2 .
y=
The cross section of
S
is a square, so we will use the formula for the area of a square:
is the side length of the square.
Since the base of
S
is a circle of radius
r,
A = s2 ,
where
s
the side length of the square is
given by the
√ r and the bottom half. The
√ top half is √
y = − r2 − x2 . Therefore, s (x) = r2 − x2 − − r2 − x2 =
√
√
2
2 r2 − x2 . Therefore, A = 2 r2 − x2 = 4 r2 − x2 . Finally, since the circle with the equations given is
centered at the origin, the x-values range from −r to r. Therefore, the following gives the volume:
the distance between the top half of a circle of radius
equation
y=
√
r 2 − x2
and the bottom half is
ˆr
4 r2 − x2 dx.
V =
−r
1