Calculus 1
HW#27 Solutions
Page 1 of 10
Section 5.1: #3, 5, 9, 13, 19, 21, 23, and 47.
Section 5.2: #5, 7, and 11.
3) Find the area between the curves y  x3 and y  x 2  1 on the interval 1  x  3.
3
A    f upper  x   flower  x   dx
1
3


  x3   x 2  1 dx
1
3
   x3  x 2  1 dx
1
3
1
1

  x 4  x3  x 
3
4
1
 81
 1 1 
   9  3     1
4
 4 3 
A  13
1
3
Check: fnInt(X3-X2+1,X,1,3)  13.333 333 33 
5) Sketch and find the area of the region determined by the intersection of the curves y  x 2  1 and y  7  x2 .
2
A
f
upper
 x   flower  x   dx
2
  7  x    x
2

2
2
2

 1 dx
2

 8  2 x  dx
2
2
2
Intersection points:
x2 1  7  x2
2x2  8
x2  4
x  2
 2  2  4  x 2  dx
(by symmetry)
0
2
1 

 4  4 x  x3 
3 0

 8
 4 8    4  0
 3
A  21
1
3
Check: fnInt(7-X2-(X2-1),X,-2,2)  21.333 333 33 
Page 2 of 10
Calculus 1
HW#27 Solutions
Section 5.1: #3, 5, 9, 13, 19, 21, 23, and 47.
Section 5.2: #5, 7, and 11.
9) Sketch and find the area of the region determined by the intersection of the curves y  4 xe x and y  x .
2
2ln  2 
A
 f
upper
 x   flower  x   dx
0
2ln  2 


2
 4 xe
 x2

 x dx
2
1 

  2e  x  x 2 
2 0



  2e



  2e

Intersection points:
4 xe
 x2
x
4 xe  x  x  0
2

x 4e
 x2
x0

1  0
OR
4e
 x2
1  0
e x 
2
1
4
1
 x 2  ln    ln  22   2 ln  2 
 4
x  2 ln  2 
A  21

2
2ln  2 
 1
2ln  2 
 1
2
2ln  2 
2
2


2
2 ln  2     2e0  0 

2
2 ln  2     2e0  0 


1
3
Check: fnInt(4Xe^(-X2)7-X,X,0,sqrt(2ln(2)))  21.333
333 33 
Calculus 1
HW#27 Solutions
Page 3 of 10
Section 5.1: #3, 5, 9, 13, 19, 21, 23, and 47.
Section 5.2: #5, 7, and 11.
13) Sketch and find the area of the region determined by the intersection of the curves y  e x and y  1  x 2 .
0
A

f
upper
 x   flower  x   dx
0.714 556 384 7
0


1  x
2
 e x  dx
0.714 556 384 7
0
 1

  x  x3  e x 
 3
 0.714 556 384 7
 0  0  e0 
Intersection points:
By inspection, (0,1) is an intersection point
1  x2  ex
1  x2  ex  0
Using Newton’s method or graphing
technology, we can find that
x  -0.714 556 384 7
1
3


  0.714 556 384 7   0.714 556 384 7   e 0.714 556 384 7 
3


A  0.082 350 246 0
Check: fnInt(1-X2-e^X,X,- 0.714 556 384 7,0) 
0.082 350 246 0 
Calculus 1
HW#27 Solutions
Page 4 of 10
Section 5.1: #3, 5, 9, 13, 19, 21, 23, and 47.
Section 5.2: #5, 7, and 11.
19) Sketch and find the area of the region determined by the intersection of the curves y  x , y  2  x , and
y  0 . Write the area as a single integral.
1
A    fupper  y   flower  y   dy
0
1
   2  y  y  dy
0
1
   2  2 y  dy
0
  2 y  y 2 
Intersection points:
By inspection, (0, 0), (1,1), and 2, 0) are the
intersection points.
This can only be written as a single integral if we take
thin horizontal slices, which implies an integral in the
y variable.
1
0
  2  1   0
A 1
Check: A = 0.5bh = 0.5(1)(2) = 1 
Calculus 1
HW#27 Solutions
Page 5 of 10
Section 5.1: #3, 5, 9, 13, 19, 21, 23, and 47.
Section 5.2: #5, 7, and 11.
21) Sketch and find the area of the region determined by the intersection of the curves x  y , x   y , and
x  1 . Write the area as a single integral.
1
A    fupper  x   flower  x   dx
0
1
   x    x   dx
0
1
   2 x  dx
0
  x 2 
Intersection points:
By inspection, (0, 0), (1,1) and (1, -1) are intersection
points
1
0
 1 0
A 1
Check: A = 0.5bh = 0.5(2)(1) = 1 
23) Sketch and find the area of the region determined by the intersection of the curves y  x , y  2 , y  6  x ,
and y = 0. Write the area as a single integral.
2
A    fupper  y   flower  y   dy
0
2
   6  x  x  dx
0
2
   6  2 x  dx
0
 6 x  x 2 
This can only be written as a single integral if we take
thin horizontal slices, which implies an integral in the
y variable.
2
0
 12  4   0  0
A8
Check: A = 0.5(b1 + b2)h = 0.5(6 + 2)(2) = 8 
Calculus 1
HW#27 Solutions
Page 6 of 10
Section 5.1: #3, 5, 9, 13, 19, 21, 23, and 47.
Section 5.2: #5, 7, and 11.
47) Consider two parabolas, each of which has its vertex at x = 0, but with different concavities. Let h be the
difference in the y-coordinates of the vertices, and w be the difference in the x-coordinates of the intersection
2
points. Show that the area between the curves is hw .
3
Let the lower curve be described by y  flower  x   a1 x 2  b1 , where both a1 > 0 and b1 > 0.
Let the upper curve be described by y  fupper  x   a2 x2  b2 , where both a2 > 0 and b2 > 0.
Notice that in order for the curves to intersect, we must have b1 > b2 and a1 > a2.
Intersection points:
a1 x 2  b1  a2 x 2  b2
 a1  a2  x 2  b1  b2
x2 
b1  b2
a1  a2
x
b1  b2
a1  a2
Notice that the intersection points can be written in terms of w:
b b 
b b 
w  1 2    1 2 
a1  a2  a1  a2 
w2

b1  b2
a1  a2
b1  b2 w

a1  a2 2
Also notice: h  b1  b2
Calculus 1
HW#27 Solutions
Section 5.1: #3, 5, 9, 13, 19, 21, 23, and 47.
Section 5.2: #5, 7, and 11.
b1 b2
a1  a2
A
 a x
2
2

 b2   a1 x 2  b1 x  dy
b b
 1 2
a1  a2
b1 b2
a1  a2
2
    a  a  x   b  b   dx
2
1
2
1
2
0
b1 b2
a1  a2
2
    a  a  x   b  b   dx
2
1
2
1
2
0
b1 b2
 a  a 
 a1  a2
 2   1 2 x 3   b1  b2  x 
3

0
  a  a   b  b 3
b b
 2   1 2  1 2    b1  b2  1 2
3
a1  a2

 a1  a2 

 a  a  b  b
 b b
 2   1 2 1 2   b1  b2   1 2
3
a1  a2

 a1  a2
 1
w
 2    b1  b2    b1  b2  
 3
2
 1

  h  h w
 3

2
 hw
3




Page 7 of 10
Calculus 1
HW#27 Solutions
Page 8 of 10
Section 5.1: #3, 5, 9, 13, 19, 21, 23, and 47.
Section 5.2: #5, 7, and 11.
x2
5) The outline of a dome is given by y  60  for -60  x  60 (units of feet), with circular cross-sections
60
perpendicular to the y-axis. Find its volume.
Since the discs are stacked up in the y direction, we need a radius that is a function of y: r(y).
And since the radius is given by the x-coordinate of a point on the curve, we need to solve the equation for the
curve for x in terms of y.
x2
y  60 
60
2
x
 60  y
60
x 2  3600  60 y
x   3600  60 y
 r  y   3600  60 y
V     r  y   dy
2
60
 

3600  60 y

2
dy
0
60
    3600  60 y  dy
0
60
  3600 y  30 y 2 
0
  3600  60  30  602   0
V  108, 000 ft 3
Calculus 1
HW#27 Solutions
Page 9 of 10
Section 5.1: #3, 5, 9, 13, 19, 21, 23, and 47.
Section 5.2: #5, 7, and 11.
7) Find the volume of a square-based pyramid with a base side length of 750 feet and a height of 500 feet.
Each slice (parallel to the xy plane) is a square whose side length decreases linearly from 750 feet to 0 feet:
750
sz  
z  750
500
3
s  z    z  750
2
V   A  z  dz
500
V

0
2
 3

  z  750  dz
 2

3
Let u   z  750
2
du
3
2

   dz   du
dz
2
3
 u  0   750
 u  750   0
V   A  z  dz
0
V
u
2
750

2
3
 2 
  du 
 3 
750
 u du
2
0
2 1 750
  u3
3 3 0
2
  7503  0
9
V  93, 750, 000 ft 3
Calculus 1
HW#27 Solutions
Page 10 of 10
Section 5.1: #3, 5, 9, 13, 19, 21, 23, and 47.
Section 5.2: #5, 7, and 11.
11) A pottery jar has a circular cross section of radius 4 + sin(x/2) inches for 0  x  2. TSketch a picture of
the jar and compute its volume.
Since the discs are stacked up in the x direction, the radius is a function of x: r(x) = 4 + sin(x/2).
V     r  x   dx
2
2
2

 x 
    4  sin    dx
 2 
0 
  fnInt((4+sin(X/2)) 2 ,X,0,2π)
 135.672 557 568