2. The Volume of the Intersection of Two Pipes
Find the volume of the
intersection of two pipes with
radius r that are joint at a
right angle as shown.
3. The Volume of the Intersection of Two Pipes
Find the volume of the
intersection of two pipes with
radius r that are joint at a
right angle as shown.
Intersection of two pipes
4. The Volume of the Intersection of Two Pipes
Find the volume of the
intersection of two pipes with
radius r that are joint at a
right angle as shown.
We’ll utilizes Cavaliere’s Intersection of two pipes
Principle which states that
the volume of a solid is the
integral of the cross-sectional area function A(x)
5. The Volume of the Intersection of Two Pipes
Find the volume of the
intersection of two pipes with
radius r that are joint at a
right angle as shown.
We’ll utilizes Cavaliere’s Intersection of two pipes
Principle which states that
the volume of a solid is the
integral of the cross-sectional area function A(x), i.e.
b b
Volume = ∫ cross-sectional areas dx = ∫ A(x) dx.
x=a x=a
6. The Volume of the Intersection of Two Pipes
Find the volume of the
intersection of two pipes with
radius r that are joint at a
right angle as shown.
We’ll utilizes Cavaliere’s Intersection of two pipes
Principle which states that
the volume of a solid is the
integral of the cross-sectional area function A(x), i.e.
b b
Volume = ∫ cross-sectional areas dx = ∫ A(x) dx.
x=a x=a
Here is a better view at the solid and it’s cross–
sections in question (see pipesvol.svr).
7. The Volume of the Intersection of Two Pipes
Find the volume of the
intersection of two pipes with
radius r that are joint at a
right angle as shown.
We’ll utilizes Cavaliere’s
Intersection of two pipes
Principle which states that
the volume of a solid is the
integral of the cross-sectional area function A(x), i.e.
b b
Volume = ∫ cross-sectional areas dx = ∫ A(x) dx.
x=a x=a
Here is a better view at the solid and it’s cross–
sections in question (see pipesvol.svr). We see that
the cross–sectional areas are squares.
8. The Volume of the Intersection of Two Pipes
We will find the volume of the
top half and double it for the
final answer.
9. The Volume of the Intersection of Two Pipes
We will find the volume of the
top half and double it for the
final answer.
As we observed, the horizontal
cross–sections are squares
10. The Volume of the Intersection of Two Pipes
We will find the volume of the
top half and double it for the
s
final answer.
As we observed, the horizontal
cross–sections are squares whose
sides s are cross–sectional lengths
of the circle.
s
11. The Volume of the Intersection of Two Pipes
We will find the volume of the
top half and double it for the
s
final answer.
As we observed, the horizontal
cross–sections are squares whose
sides s are cross–sectional lengths
of the circle. Let x and r be as
shown.
s
r x
12. The Volume of the Intersection of Two Pipes
We will find the volume of the
top half and double it for the
s
final answer. r x
As we observed, the horizontal
cross–sections are squares whose
sides s are cross–sectional lengths
of the circle. Let x and r be as
shown.
s
r x
13. The Volume of the Intersection of Two Pipes
We will find the volume of the
top half and double it for the
s
final answer. r x
As we observed, the horizontal
cross–sections are squares whose
sides s are cross–sectional lengths
of the circle. Let x and r be as
shown. Hence
s
s/2 = √r2 – x2
r x
14. The Volume of the Intersection of Two Pipes
We will find the volume of the
top half and double it for the
s
final answer. r x
As we observed, the horizontal
cross–sections are squares whose
sides s are cross–sectional lengths
of the circle. Let x and r be as
shown. Hence
s
s/2 = √r2 – x2 or that
r x
s = 2√r2 – x2
where x goes from 0 to r.
15. The Volume of the Intersection of Two Pipes
We will find the volume of the
top half and double it for the
s
final answer. r x
As we observed, the horizontal
cross–sections are squares whose
A(x) = 4(r2 – x2)
sides s are cross–sectional lengths
of the circle. Let x and r be as
shown. Hence
s
s/2 = √r2 – x2 or that
r x
s = 2√r2 – x2
where x goes from 0 to r.
Therefore the cross–sectional area function is
A(x) = s2 = 4(r2 – x2).
16. The Volume of the Intersection of Two Pipes
So the volume of the top is
r
∫0 4(r2 – x2) dx s
r s
x
A(x) = 4(r2 – x2)
s = 2√r2 – x2
b r x
x=a
17. The Volume of the Intersection of Two Pipes
So the volume of the top is
r
∫0 4(r2 – x2) dx s
r s
r x
= 4∫ (r2 – x2) dx
0
A(x) = 4(r2 – x2)
s = 2√r2 – x2
b r x
x=a
18. The Volume of the Intersection of Two Pipes
So the volume of the top is
r
∫0 4(r2 – x2) dx s
r s
r x
= 4∫ (r2 – x2) dx
0
r
= 4(r x – x /3) | x=0
2 3 A(x) = 4(r2 – x2)
s = 2√r2 – x2
b r x
x=a
19. The Volume of the Intersection of Two Pipes
So the volume of the top is
r
∫0 4(r2 – x2) dx s
r s
r x
= 4∫ (r2 – x2) dx
0
r
= 4(r x – x /3) | x=0
2 3 A(x) = 4(r2 – x2)
= 4(2r3/3) = 8r3/3
s = 2√r2 – x2
b r x
x=a
20. The Volume of the Intersection of Two Pipes
So the volume of the top is
r
∫0 4(r2 – x2) dx s
r s
r x
= 4∫ (r2 – x2) dx
0
r
= 4(r x – x /3) | x=0
2 3 A(x) = 4(r2 – x2)
= 4(2r3/3) = 8r3/3
s = 2√r2 – x2
Therefore the entire volume
of the intersection is twice b r x
the above or 16r3/3. x=a
