# PRACTICE PROBLEMS: SET 4 1. Problems x y

```PRACTICE PROBLEMS: SET 4
MATH 101: PROF. DRAGOS GHIOCA
1. Problems
Problem 1. Compute the area of the region bounded by y = cos(x), y = cos(2x),
x = π2 and x = π.
Problem 2. Find the volume of the solid obtained by rotating the region between
the curves y = x4 and x = y 4 around the line y = 5.
Problem 3. Find a positive number c such that the region between the curve
y = c − x2 and the x-axis has area equal to 5.
1
2
MATH 101: PROF. DRAGOS GHIOCA
2. Solutions
Problem 1. The area of the region between the graphs of cos(x) and of cos(2x)
bounded by x = π2 and x = π is
∫ π
|cos(x) − cos(2x)| dx .
π
2
So, we need to determine which function (cos(x) or cos(2x)) is larger on each
subinterval of [π/2, π]. So, for this we solve what are the points of intersection for
the two graphs. Therefore, we equate:
cos(x) = cos(2x).
In order to solve the above equation we note the identity:
cos(2x) = 2 cos2 (x) − 1,
and thus we have to solve:
cos(x) = 2 cos2 (x) − 1,
which yields 2 cos2 (x) − cos(x) − 1 = 0, and hence
(2 cos(x) + 1)(cos(x) − 1) = 0.
[
]
So, we obtain either cos(x) = − 12 or cos(x) = 1. We note that on the interval π2 , π
the cosine only takes values less than or equal to 0, and thus we only have to solve
1
cos(x) = − .
2
[
]
The unique solution x ∈ π2 , π is x = 2π
.
More
precisely,
3
(
)
1
π 2π
cos(x) &gt; − for x ∈
,
2
2 3
and therefore (since always cos(x) − 1 &lt; 0), we get
(
(2 cos(x) + 1)(cos(x) − 1) &lt; 0 for x ∈
Similarly,
π 2π
,
2 3
(
(2 cos(x) + 1)(cos(x) − 1) &gt; 0 for x ∈
which in particular yields that
(
cos(x) &lt; cos(2x) for x ∈
and cos(x) &gt; cos(2x) for x ∈
∫ π
|cos(x) − cos(2x)| dx
π
2
(π
2π
2, 3
∫
2π
3
=
π
2
2π
,π
3
)
.
)
2π
,π ,
3
)
)
. Therefore the desired area equals
∫
cos(x) − cos(2x) dx +
π
2π
3
cos(2x) − cos(x) dx .
In order to evaluate (using the Fundamental Theorem of Calculus) the above definite integrals we need to find the antiderivative of cos(2x). We do this using the
substitution
u = 2x and so, du = 2 dx .
PRACTICE PROBLEMS: SET 4
Thus
∫
∫
3
cos(u)
sin(u)
sin(2x)
du =
+C =
+C.
2
2
2
cos(2x) dx =
We finally compute
∫ π
|cos(x) − cos(2x)| dx
π
2
(
(
)
)
2π
sin(2x)
sin(2x)
sin(x) −
− sin(x) |π2π
| π3 +
2
3
2
2
( √
(√
√ )
√ )
3
3
3
3
+
− (1 − 0) + (0 − 0) − −
−
2
4
4
2
√
3 3
−1
2
√
3 3−2
.
2
=
=
=
=
Problem 2. We first find the points where those two curves intersect. So, we
have
x = y 4 = (x4 )4 = x16 .
From the equation x = x16 we obtain that either x = 0 or x15 = 1 in which case
we have x = 1. So, the two curves intersect in two points: (0, 0) and (1, 1).
√
We note that for x ∈ [0, 1], the curve x√= y 4 is the same with the curve y = 4 x.
Moreover, on this interval, we have that 4 x ≥ x4 .
When we rotate the region around the line y = 5, we obtain the general sections,
when we slice the region at a point x between 0 and 1, are annuli (i.e., region
between two circles). The radius of the large circle is the distance between the
curve y = x4 and the line y = 5, and thus, it is 5 − x4 . On the √
other hand, the
4
is
the
distance
between
the
curve
y
=
x and the line
√
y = 5, and so, it is 5 − 4 x. In conclusion, the volume of the solid is
∫ 1
(
√ )
π &middot; (5 − x4 )2 − (5 − 4 x)2 dx
0
∫
1
=
0
(
√
√ )
π &middot; 25 − 10x4 + x8 − 25 + 10 4 x − x dx
∫
(
)
1
1
π &middot; x8 − 10x4 − x 2 + 10x 4 dx
0
)
(
3
5
x9
x2
x4
5
=π&middot;
− 2x − 3 + 10 &middot; 5
|10
9
2
4
)
(
2
1
−2− +8
=π&middot;
9
3
49π
=
.
9
1
=
Problem 3. The parabola
− x2 is symmetric with respect to y-axis. More√y = c √
over, its x-intercepts are − c and c (note that it is given to us that c is positive).
4
MATH 101: PROF. DRAGOS GHIOCA
So, the area of the region is double the area enclosed by the x-axis, y-axis and the
curve y = c − x2 . In other words, the area equals
∫ √c
2&middot;
c − x2 dx
0
(
)
x3 √c
= 2 &middot; cx −
|0
3
(
√ )
√
( c)3
=2&middot; c&middot; c−
3
3
=
4c 2
.
3
So, we need to find c such that
3
4c 2
= 5,
3
3
which yields c 2 =
15
4
and so,
(
c=
15
4
) 23
.
```