MATH 115, T2. Solutions for assignment 2.
1. (10pt)
Z
sin x
dx =
cos3 x
substitution u = cos x, du = − sin xdx,
Z
=−
du
1
1
1
= u−2 + C =
+ C = sec2 x + C.
3
2
u
2
2 cos x
2
2. (10pt) Find the area between f (x) = sin x and g(x) = sin 2x from x = 0 to x = π/2.
The plots of sin x and sin 2x intersect for x satisfying sin x = sin 2x = 2 sin x cos x, sin x (1 − 2 cos x) =
0, that is for sin x = 0 or cos x = 1/2. On the given interval the intersection points are x = 0
and x = π/3, for 0 < x < π/3 sin 2x > sin x, for π/3 < x < π/2 sin 2x < sin x. The area is
equal to
A=
Z π/2
0
|sin 2x − sin x| dx =
Z π/3
0
(sin 2x − sin x) dx +
Z π/2
π/3
(sin x − sin 2x) dx
For the integrals of sin 2x we use substitution u = 2x, dx = 21 du,
Z π/3
0
sin 2xdx =
1
1 Z 2π/3
sin udu = − cos u
2 0
2
¸2π/3
µ
=−
0
¶
µ
¶
1
1
1
3
−
− −
= ,
2
2
2
4
Z π/3
1
1
sin xdx = − cos x]π/3
=− +1= ,
0
2
2
0
Z π/2
1
1
π/2
sin xdx = − cos x]π/3 = 0 + = ,
2
2
π/3
Z π/2
π/3
1
sin 2xdx = − cos u
2
so
A=
¸π
2π/3
µ ¶
1
1
1
= − (−1) −
= ,
2
4
4
3 1 1 1
1
− + − = .
4 2 2 4
2
3. (10 pt) Find the area between y = cos x, y = sec2 x from x = 0 to x = π/4.
The only intersection point is x = 0, for 0 < x ≤ π/4 sec x > 1 while cos x < 1. Therefore
√
Z π/4 ³
´
2−1
1
π/4
A=
sec2 x − cos x dx = (tan x − sin x)]0 = 1 − √ = √ .
0
2
2
4. (10pt) Find the volume of the solid obtained by rotating the region bounded by x = y − y 2 ,
x = 0, about y-axis.
The region corresponds to x ≥ 0, 0 ≤ y ≤ 1 (see picture). For the given y the section of the
solid by the plane orthogonal to y-axis is a disk of radius R = x = y − y 2 = y (1 − y), its area
is A(y) = πR2 = πy 2 (1 − y)2 , so the volume is
V =
=π
Z 1³
0
Z 1
0
2
A(y)dy = π
3
y − 2y + y
4
´
Z 1
0
2
2
y (1 − y) dy = π
Z 1
0
µ
1 3 1 4 1 5
dy = π
y − y + y
3
2
5
³
´
y 2 1 − 2y + y 2 dy =
¶¸1
0
µ
¶
π
1 1 1
− +
= .
=π
3 2 5
30
5. (10pt) Find the volume of the solid obtained by rotating the region bounded by y = x2 ,
y = 4, about the line y = 4.
The region corresponds to −2 ≤ x ≤ 2, 0 ≤ y ≤ 4 (see picture). For the given x the section
of the solid by the plane orthogonal to x-axis is a disk of radius R = 4 − x2 , its area is
2
A(x) = πR2 = π (4 − x2 ) , so the volume is
V =
Z 2
−2
A(x)dx = π
Z 2 ³
−2
4−x
´
2 2
dx = π
Z 2 ³
−2
´
16 − 8x2 + x4 dx =
integral of an even function in symmetric limits
= 2π
Z 2³
0
2
16 − 8x + x
4
´
µ
8
1
dx = 2π 16x − x3 + x5
3
5
¶¸2
0
µ
¶
64 32
512
= 2π 32 −
+
=
π.
3
5
15
6. (20pt) Find the volume of the solid obtained by rotating the region bounded by x + 3y = 2,
x = 0, y = 0 about 1) x-axis 2) y-axis.
The domain is a triangle shown in the picture.
a) Cross-section orthogonal to the axis of rotation for a given x is a disk of the radius R = y =
1
(2 − x), its area is A(x) = πR2 = π9 (2 − x)2 , so the volume is
3
V =
Z 2
0
´
πZ2
π Z 2³
2
A(x)dx =
(2 − x) dx =
4 − 4x + x2 dx =
9 0
9 0
µ
¶¸
µ
¶
2
π
1
8
8π
π
4x − 2x2 + x3
8−8+
=
.
=
9
3
9
3
27
0
b) Cross-section orthogonal to the axis of rotation for a given y is a disk of the radius R = x =
(2 − 3y), its area is A(y) = πR2 = π (2 − 3y)2 , so the volume is
=
V =
Z 2/3
0
A(y)dy = π
³
Z 2/3
0
2
2
(2 − 3y) dy = π
= π 4y − 6y + 3y
3
´i2/3
0
µ
Z 2/3 ³
0
´
4 − 12y + 9y 2 dy =
¶
8 8 8
8π
− +
=
.
=π
3 3 9
9
7. (10pt) Set up, but do not evaluate, an integral for the volume of the solid obtained by
rotating the region bounded by 2x + 3y = 6, (y − 1)2 = 4 − x, about the line x = −5.
The region is shown in the figure. The points of intersection
³
´of the line x = g(y) = (6 − 3y) /2
2
9 7
and the parabola x = f (y) = 4−(y − 1) are (3, 0) and − 4 , 2 , this gives the limits of y change.
Cross-section at the given y is a washer with the internal radius r = g(y) − (−5) = g(y) + 5,
the external radius R = f (y) + 5, and the area A(y) = π (R2 − r2 ). So the volume is
V =
Z 7/2
0
A(y)dy = π
Z 7/2 ó
0
9 − (y − 1)
´
2 2
µ
3
− 8− y
2
¶2 !
dy.
8. (20pt) Base of a solid S is a circular disk with radius r. Parallel cross sections perpendicular
to the base are squares. Find the volume of the solid.
This problem is close to Example 7, section 6.2 in the textbook, see the corresponding figures,
but instead of triangles we have squares. So let the equation for the circle in the base is
x2 +√y 2 = r2 , let the squares be orthogonal to x-axis, then the side of a square is a = 2y,
y = r2 − x2 , −r ≤ x ≤ r. The area of the square is A(x) = a2 = 4 (r2 − x2 ), and the volume
is
¸
Z r ³
´
4 3 r
16
2
2
2
V =4
r − x dx = 4r x − x
= r3
3 −r
3
−r