Section 6.2: Volume
Consider a solid whose base is the region bounded by
y=1-x, y=2x+5, x=0 and x=3.
If the cross sections are squares,
set up an integral to find volume.
Volume of box = height * width * length
Consider a solid whose base is the region bounded by
y=1-x, y=2x+5, x=0 and x=3.
This isn't a box,
but we could break it into pieces,
pretend they are boxes,
then take limit.
height = (2x+5) - (1-x)
length = (2x+5) - (1-x)
width = x

Volume    2 x  5   1  x
n
k 1
*
k
*
k

2
xk
3
Volume     2 x  5   1  x   dx  237
2
0
What if cross sections are circles??
  2 x  5   (1  x) 
 x
 
2


2
  2 x  5   (1  x) 

dx


0 
2

3
2
Consider region bounded by y=x 2 and y=2x
Find volume when rotated about:
x-axis (using washers)
If we slice perpendicular to x-axis, we get a washer
Thickness=x
2
2 2

(2
x
)


(
x
)  x

2

(2
x
)


(
x
)
dx

2
0
2 2
y = -3
2
  (2 x  3)
2
  ( x  3) dx
2
2
0
Important:
The axis is the center of the washer
5
y=5
2
  (5  x
)   (5  2 x) dx
2 2
2
0
Important: The axis is the center of the washer
Remember that we also treated x as a function of y.
1
x= y
and x= y
2
Rotate about the x-axis
Cut Parallel to the axis
We get a shell with thickness y
1
y y
2
Unroll the shell
1 

0 2 y  y  2 y  dy
4
2Πy
y = -3 (shells)
1
y y
2
2Π(y-(-3))
1 

0 2 ( y  3)  y  2 y  dy
4
y = 5 (shells)
1
y y
2
2Π(5-y)
1

2

(5

y
)
y


0
2

4

y  dy

We can rotate about the y-axis as well
y = x2
y = 2x
x = √y
x=½y
y-axis
½y
√y
4


0
 y
2
2
1 
   y  dy
2 
We can rotate about the y-axis as well
y-axis
2Πx
2x  x 2
2x
x2
x
|
2
2

x
(2
x

x



0
2
) dx
x  3
2Π (x+3)
½y
√y
2
2x  x 2
(2
x

x
)
2

(
x

3)
dx



2
0
x  3
½y
√y
4
1
2
0  ( y  3)   ( 2 y  3) dy
2
x=5
½y
√y
2

1 

0   5  2 y    5  y
4

2
dy
x=5
2x
x2
2Π (5 - x)
2x  x 2
2
  2 (5  x) 
0
(2 x  x ) dx
2