Fluid is flowing in a tube that has a radius of 3 centimeters. Water is flowing through a circular cross section at a rate of (9-r^2) cm/s, where r is the distance from the center of the cross section. What is the total amount of water that flows through the cross section in 4 seconds?
at a radius r, the volume flowing through per second is
dv = 2πrh dr
where h is the height of the cylinder. That height is the rate of flow per second, so
dv = 2πr(9-r^2) dr
dv/dr = 2πr(9-r^2)
Now add up for all the thin cylinders of thickness dr, and
v = ∫[0,3] 2πr(9-r^2) dr = 81π/2 cm^3
This problem seems kind of hard to understand, but consider a stack of nested cylinders of varying heights, where the inner cylinders are taller than the outer ones, where the water flows more slowly. The area of a circular ring of radius r and width dr is just 2πr dr.
To find the total amount of water that flows through the cross section in 4 seconds, we need to calculate the volumetric flow rate first.
The volumetric flow rate is the product of the cross-sectional area and the velocity of the fluid at that cross section.
The cross-sectional area of a circle is given by the formula:
A = πr^2
Here, r is the radius of the tube, which is 3 centimeters.
A = π × (3 cm)^2
A = π × 9 cm^2
A = 9π cm^2
Now, let's calculate the velocity of the fluid at each distance from the center of the cross section.
Given that the velocity of the fluid is (9 - r^2) cm/s, we can substitute r = 3 cm to find the velocity at the periphery of the cross section.
v = (9 - r^2)
v = (9 - (3 cm)^2)
v = (9 - 9 cm^2)
v = 0 cm/s
So, the velocity at the periphery of the cross section is 0 cm/s.
Now, let's calculate the volumetric flow rate:
Q = Av
Q = (9π cm^2) × (0 cm/s)
Q = 0 cm^3/s
Therefore, the total amount of water that flows through the cross section in 4 seconds would be:
Total amount = Q × time
Total amount = (0 cm^3/s) × (4 s)
Total amount = 0 cm^3
Hence, the total amount of water that flows through the cross section in 4 seconds is 0 cubic centimeters.
To find the total amount of water that flows through the cross-section in 4 seconds, we need to calculate the volume of fluid that passes through the cross-section per second and then multiply it by the duration of time.
First, let's find the rate of fluid flow at each point in the cross-section. The rate of flow at any point is given by (9-r^2) cm/s, where r is the distance from the center of the cross-section.
To determine the total volume of fluid that passes through the cross-section per second, we need to integrate the rate of flow function over the cross-sectional area.
The cross-sectional area of a circular tube is given by A = πr^2, where r is the radius of the tube. In this case, the radius is 3 centimeters, so the cross-sectional area is A = π(3)^2 = 9π cm^2.
Now, let's integrate the rate of flow function (9-r^2) over the cross-sectional area:
∫(9-r^2) dA,
where dA is the differential area element in terms of r.
∫(9-r^2) dA = ∫(9-r^2) 2πr dr,
where the limits of integration are from 0 to 3 (since the radius of the tube is 3 cm).
∫(9-r^2) 2πr dr = 2π∫(9r - r^3) dr = 2π[(4.5r^2) - (0.25r^4)] evaluated from 0 to 3,
= 2π[(4.5(3)^2) - (0.25(3)^4)] - 2π[(4.5(0)^2) - (0.25(0)^4)],
= 2π[(4.5(9) - 0.25(81)] - 0,
= 2π[(40.5 - 20.25)].
Calculating further, we get:
= 2π[20.25],
= 40.5π cm^3/s.
Now, to find the total amount of water that flows through the cross-section in 4 seconds, we need to multiply the rate of flow by the duration of time:
Total amount of water = (40.5π cm^3/s) * (4 s),
= 162π cm^3.
Therefore, the total amount of water that flows through the cross-section in 4 seconds is 162π cm^3.