Hellen SUHADI
Assignment Homework-1 due 09/15/2021 at 11:59pm HKT
1. (1 point)
Sketch the region enclosed by the curves and find its area.
y = x, y = 4x, y = −x + 2
AREA =
2021 FALL MATH1014
Area =
Answer(s) submitted:
• 6-(3/2)*pi
(correct)
Answer(s) submitted:
• 0.6
5. (1 point)
Sketch the region enclosed by the curves and find its area.
y = ex , y = e3x , x = ln 2, x = ln 4
AREA =
(correct)
2. (1 point) Sketch the region enclosed by the given curves.
Decide whether to integrate with respect to x or y. Then find the
area of the region.
y = 6x2 , y = x2 + 6
Answer(s) submitted:
• 56/3 -2
(correct)
6. (1 point) Consider the area between the graphs x + 3y = 14
and x + 4 = y2 . This area can be computed in two different ways
using integrals
Answer(s) submitted:
• (8*30ˆ(1/2))/5
First of all it can be computed as a sum of two integrals
(correct)
Z b
Z c
f (x) dx +
a
where a =
f (x) =
g(x) =
g(x) dx
b
,b=
,c=
and
Alternatively this area can be computed as a single integral
Z β
h(y) dy
α
where α =
,β=
h(y) =
Either way we find that the area is
and
.
Answer(s) submitted:
•
•
•
•
•
•
•
•
•
3. (1 point)
Find the area of the region enclosed between f (x) = x2 +
5x + 23 and g(x) = 2x2 + 3x − 1.
Area =
(Note: The graph above represents both functions f and g
but is intentionally left unlabeled.)
Answer(s) submitted:
• 500/3
-4
5
32
2((x+4)ˆ(1/2))
-x/3+14/3+(x+4)ˆ(1/2)
-6
3
-yˆ2-3y+18
243/2
(correct)
(correct)
7. (1 point) Find the value(s) of c such that the area of the
region bounded by the parabolae y = x2 − c2 and y = c2 − x2 is
72.
Answer (separate by commas): c =
4. (1 point) Sketch the region enclosed by the curves given
below. Decide whether to integrate with respect to x or y. Then
find the area of the region.
6x
y = 3 cos x, y = 3 − .
π
Answer(s) submitted:
• 3, -3
1
(correct)
the slice used to find the integral, labeling the variable and differential on your sketch. Then evaluate the integral to find the
area. R
A. 06 π(3 − y/2)2 dy
Which is the shape of the region being integrated?
8. (1 point)
Find the area of the shaded region below.
• A. Cone
• B. Hemisphere
radius/radius and height =
(Enter the radius, or the radius and height separated by a
comma, e.g., 4, 3)
volume
=
R
B. 013 π(169 − h2 ) dh
Which is the shape of the region being integrated?
Area =
Answer(s) submitted:
• 9
(correct)
• A. Cone
• B. Hemisphere
9. (1 point) The region bounded by y = ex , y = 0, x = −1, x =
0 is rotated around the x-axis. Find the volume.
volume =
radius/radius and height =
(Enter the radius, or the radius and height separated by a
comma, e.g., 4, 3)
volume =
Answer(s) submitted:
• pi(0.5-(1/(2*eˆ2)))
(correct)
Answer(s) submitted:
10. (1 point)
Find the volume of the solid whose base is the circle x2 + y2 = 9
and the cross sections perpendicular to the x-axis are triangles
whose height and base are equal.
Find the area of the vertical cross section A at the level x = 2.
•
•
•
•
•
•
A
3, 6
18pi
B
13
4394pi/3
A=
(correct)
V=
13. (1 point) Find the volume of the solid S described below.
Answer(s) submitted:
The base of S is the region enclosed by the parabola y = 1 − x2
and the x-axis. Cross-sections perpendicular to the y-axis are
squares.
• 10
• 72
(correct)
Volume =
Answer(s) submitted:
11. (1 point)
Find the volume of the solid whose base is the region enclosed
by y = x2 and y = 3, and the cross sections perpendicular to the
y-axis are squares.
V=
• 2
(correct)
14. (1 point)
Use calculus to find the volume of the following solid S:
The base of S is the parabolic region {(x, y)| x2 ≤ y ≤ 1}.
Cross-sections perpendicular to the y-axis are equilateral triangles.
Volume =
Answer(s) submitted:
• 18
(correct)
12. (1 point) Each of the following integrals represents the
volume of either a hemisphere or a cone, and the variable of
integration measures a length. In each case, say which shape
is represented and give the radius of the hemisphere or radius
and height of the cone. Make a sketch of the region, showing
Answer(s) submitted:
• sqrt(3)/2
(correct)
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