Intersection of Curves and Calculation of Areas and Volumes, Assignments of Calculus

Solutions to mathematics exercises from millersville university. The first exercise involves finding the area between the curves y = x^3 - 2x^2 and y = x^2 by integrating the difference of their functions. The second exercise calculates the volume of a triangular prism representing the attic, given its base area and length.

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Pre 2010

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Millersville University Name Answer Key
Department of Mathematics
MATH 211, Homework 02
January 30, 2006
Page 409, Exercise 18
Sketch and find the are of the region determined by the intersection of the curves:
y=x3
โˆ’2x2, y =x2.
0.5 1 1.5 2 2.5 3 x
2
4
6
8
y
0.5 1 1.5 2 2.5 3 x
2
4
6
8
y
The curves intersect when the following equation is satisfied:
x3
โˆ’2x2=x2
x3
โˆ’3x2= 0
x2(xโˆ’3) =
x= 0 or x= 3
Thus the area between the curves is
A=Z3
0
x2
โˆ’๎˜€x3
โˆ’2x2๎˜dx
=Z3
0
โˆ’x3+ 3x2dx
=โˆ’
1
4x4+x3๎˜Œ
๎˜Œ
๎˜Œ
๎˜Œ
3
0
=27
4
pf2

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Millersville University Name Answer Key

Department of Mathematics

MATH 211, Homework 02

January 30, 2006

Page 409, Exercise 18

Sketch and find the are of the region determined by the intersection of the curves:

y = x

3 โˆ’ 2 x

2 , y = x

2 .

x

y

x

y

The curves intersect when the following equation is satisfied:

x

3 โˆ’ 2 x

2 = x

2

x

3 โˆ’ 3 x

2 = 0

x

2 (x โˆ’ 3) =

x = 0 or x = 3

Thus the area between the curves is

A =

0

x

2 โˆ’

x

3 โˆ’ 2 x

2

dx

0

โˆ’x

3

  • 3x

2 dx

x

4

  • x

3

3

0

Page 423, Exercise 14

The diagram below illustrates a triangular cross section of the attic.

height = 10 ft

base = 30 ft

The cross sectional area of the attic is the constant

A =

bh =

The length of the attic is 60 feet and thus the volume of the attic is

V =

60

0

A dx =

60

0

150 dx = (60)(150) = 9000 ft

3 .