423
Answers
601–700
Answers and Explanations
If you consider a cross-sectional slice at point (x, y) on the circle, where y > 0, the base
of the square equals 2y; therefore, the area of the cross-sectional slice is (2y)
2
= 4y
2
.
In order to integrate with respect to x, you can solve the equation of the circle for y
2
and write the area as 4y
2
= 4(16 – x
2
) = A(x). Because the base of the solid varies for x
over the interval [–4, 4], the limits of integration are –4 and 4. Therefore, the integral to
find the volume is
4 16
6 4 4
2 64 4
2 64
4
3
2
4
4
2
4
4
2
0
4
3
−
(
) =
−
(
)
=
−
(
)
=
−
−
−
∫
∫
∫
x dx
x dx
x dx
x
x
=
−
=
0
4
3
2 64 4
4 4
3
1 024
3
( )
( )
,
Answers
601–700
Answers and Explanations
If you consider a cross-sectional slice at point (x, y) on the circle, where y > 0, the base
of the square equals 2y; therefore, the area of the cross-sectional slice is (2y)
2
= 4y
2
.
In order to integrate with respect to x, you can solve the equation of the circle for y
2
and write the area as 4y
2
= 4(16 – x
2
) = A(x). Because the base of the solid varies for x
over the interval [–4, 4], the limits of integration are –4 and 4. Therefore, the integral to
find the volume is
4 16
6 4 4
2 64 4
2 64
4
3
2
4
4
2
4
4
2
0
4
3
−
(
) =
−
(
)
=
−
(
)
=
−
−
−
∫
∫
∫
x dx
x dx
x dx
x
x
=
−
=
0
4
3
2 64 4
4 4
3
1 024
3
( )
( )
,