Math 2260 HW #2 Solutions
Math 2260 Written HW #2 Solutions
1. Find the area of the region that is enclosed between the curves y = 2 sin(x) and y = 2 cos(x) from x = 0 to x = /2. Answer: The desired area is pictured below:
2
1.5
1
0.5
0
0.25
0.5
0.75
1
1.25
1.5
1.75
In the figure, the blue curve is y = 2 sin(x) and the red curve is y = 2 cos(x). Since the top of the region is the blue curve between 0 and /4 (since 2 cos(x) = 2 sin(x) when x = /4), whereas the top of the region is the red curve between /4 and /2, we need to compute the area in two parts:
Area = =
/4
/2
2 sin(x) dx +
2 cos(x) dx
0
/4
/4
/2
- 2 cos(x) + 2 sin(x)
0
/4
= [-2 cos(/4) + 2 cos(0)] + [2 sin(/2) - 2 sin(/4)]
1
1
= -2 ? + 2 ? 1 + 2 ? 1 - 2 ?
2
2
= - 2+2 + 2- 2
= 4 - 2 2.
Therefore, the area of the shaded region is 4 - 2 2.
1
2. Find the volume of the solid generated by revolving the region bounded by the x-axis and the semicircle y = 4 - x2 around the x-axis. Answer: First, here's the graph of the function y = 4 - x2:
2.4 1.6 0.8
-2.4
-1.6
-0.8
0
-0.8
0.8
1.6
2.4
-1.6
-2.4
Revolving around the x-axis will yield a sphere like this one:
Now, each cross section of the sphere is a circle of radius 4 - x2, so we know that the cross-sectional area is
A(x) = 4 - x2 2 = (4 - x2) = 4 - x2.
2
Nowthe volume of this solid is given by integrating A(x) along its length. Since the graph y = 4 - x2 intersects the x-axis at x = ?2, we should integrate from -2 to 2. Therefore, the volume of the sphere is
2
A(x) dx =
-2
=
=
2
4 - x2 dx
-2
x3 2 4x -
3 -2
23 4(2) - ? -
3
(-2)3 4(-2) - ?
3
8
8
= 8 - - -8 +
3
3
16 = 16 -
3 32 =.
3
Thus,
the
volume
of
the
given
solid
is
32 3
.
3
................
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