Water is flowing in a cylindrical pipe of radius 1 inch. Because water is viscous and sticks to the pipe, the rate of flow varies with distance from the center. The speed of the water at distance r inches from the center is 10(1-(r^2)) per second. What is the rate (in cubic inches per second) at which water is flowing through the pipe?

consider the amount of water that flows through a cross-section of the pipe in 1 second. It is a conical solid, high in the center and zero at the edges.

At radius r, the volume of the cylindrical shell of water is 2pi*r*h dr, where h = 10(1-r^2)

So, the amount across the whole cross-section is

integral[0,1] 2pi*r*10(1-r^2) dr
= 10pi r^2 - 5pi r^4 [0,1]
= 5pi in^3

since that's the volume in 1 second, the rate is 5pi in^3/s

To find the rate at which water is flowing through the pipe, we need to calculate the cross-sectional area of the pipe at each distance from the center and then multiply it by the corresponding speed of the water.

The cross-sectional area of a cylindrical pipe with radius r can be found using the formula:
A = π * r^2

Now, let's calculate the rate at which water is flowing through the pipe.

We need to integrate the product of the cross-sectional area (π * r^2) and the speed of the water (10(1-(r^2))) over the entire range of the radius (from 0 to 1) to find the total flow rate in cubic inches per second.

Rate of flow = ∫[0,1] (π * r^2) * (10(1-(r^2))) dr

Let's solve this integral step-by-step.

Step 1: Expand the expression inside the integral.
Rate of flow = ∫[0,1] (10π(r^2 - r^4)) dr

Step 2: Integrate each term separately.
Rate of flow = ∫[0,1] (10πr^2 - 10πr^4) dr

Step 3: Integrate using the power rule.
Rate of flow = (10π/3)r^3 - (10π/5)r^5 + C

Step 4: Evaluate the integral limits.
Rate of flow = [(10π/3)(1^3) - (10π/5)(1^5)] - [(10π/3)(0^3) - (10π/5)(0^5)]

Step 5: Simplify the expression.
Rate of flow = [(10π/3) - (10π/5)] - 0

Step 6: Simplify the fractions.
Rate of flow = [(50π - 30π)/15]

Step 7: Simplify the expression.
Rate of flow = (20π/15)

Step 8: Simplify the fraction.
Rate of flow = (4π/3) cubic inches per second

Therefore, the rate at which water is flowing through the pipe is (4π/3) cubic inches per second.

To find the rate at which water is flowing through the pipe, we need to calculate the volume of water passing through each infinitesimally thin cylindrical section of the pipe and then integrate those volumes for the entire length of the pipe.

Let's consider an infinitesimally thin cylindrical section of the pipe with radius r and thickness dr. The volume of water passing through this section in one second can be calculated by multiplying the cross-sectional area of the section by the speed of the water at that radius.

The cross-sectional area of the cylindrical section is given by A = πr^2 (since the pipe has a constant radius of 1 inch).

The speed of the water at distance r from the center is given by v = 10(1 - r^2) inches per second.

The volume of water passing through the infinitesimally thin cylindrical section in one second is therefore given by dV = A * v * dt = πr^2 * (10(1 - r^2)) * dt.

Integrating this expression for V, the total volume of water passing through the entire length of the pipe, we get:
V = ∫[0,1] πr^2 * (10(1 - r^2)) * dr.

To calculate this integral, expand the expression and integrate term by term:
V = ∫[0,1] (10πr^2 - 10πr^4) * dr.

Integrating term by term:
V = [10πr^3/3 - 10πr^5/5] evaluated from 0 to 1.

Plugging in the upper and lower limits:
V = (10π/3 - 10π/5) - (0) = (10π/3 - 2π) cubic inches per second.

Therefore, the rate at which water is flowing through the pipe is (10π/3 - 2π) cubic inches per second.