Part II: The Answers
426
Answers
601–700
Because the base of the solid varies for y over the interval [–2, 2] the limits of integration
are –2 and 2. Therefore, the integral to find the volume is
36 9
2 36 9
2 36
3
2 36 2 3 2
2
2
2
2
0
2
3
0
2
3
−
(
) =
−
(
)
=
−
(
)
=
−
(
)
=
−
∫
∫
y dy
y dy
y
y
( ) ( )
9 96
Note that because 36 – 9y
2
is an even function of y, the lower limit of integration was
changed to zero and the resulting integral was multiplied by 2 to make the integral a
bit easier to compute.
685.
8
3
Recall the definition of volume: Let S be a solid that lies between y = c and y = d.
If A(y) is the cross-sectional area of S in the plane P y that goes through y and is
perpendicular to the y-axis, where A is a continuous function, then the volume
of S is
V
A y dx
c
d
= ∫ ( )
Here, you want to find an expression for A(y).
The base of S is the region bounded by the curves x = 0, y = 0, and x
y
= −
2 2
. If you consider a
cross-sectional slice that goes through a point (x, y) on the line x
y
= −
2 2
, the cross-sectional
slice is an isosceles triangle with height equal to the base, so the area is
1
2
1
2
1
2
1
2
2 2
2
2
bh
x x
x
y
A y
= ( )( ) =
=
−





 = ( )