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Oil Flow Oil flows through a cylindrical pipe of radius 3 inches, but friction from the pipe slows the flow toward the outer edge. The speed at which the oil flows at a distance \(r\) inches from the center is 8\(\left(10-r^{2}\right)\) inches per second. (a) In a plane cross section of the pipe, a thin ring with thickness \(\Delta r\) at a distance \(r\) inches from the center approximates a rectangular strip when you straighten it out. What is the area of the strip (and hence the approximate area of the ring)? (b) Explain why we know that oil passes through this ring at approximately 8\(\left(10-r^{2}\right)(2 \pi r) \Delta r\) cubic inches per second. (c) Set up and evaluate a definite integral that will give the rate (in cubic inches per second) at which oil is flowing through the pipe.

Step by step solution

01

Compute the area of the ring

The area of the rectangular strip, which consists of the circumference of the ring (2蟺r) multiplied by the thickness 螖r, gives the approximate area of this ring. Using the formula for the area of a rectangle (length 脳 width), we can find the area \(A = 2蟺r 螖r\).

02

Compute the oil flow through the ring

The speed at which oil flows through this ring (or through the approximated rectangular strip) is given as \(8(10-r^{2})\). Hence, the volume flow rate (volume/time) would be the speed times the cross-section area of flow (A), which we calculated in Step 1. So, the volume flow rate \(V_{rate} = 8(10-r^{2}) 2蟺r 螖r\) cubic inches per second.

03

Setup the integral for the whole pipe

The total volume flow rate through the entire pipe would be the sum of the flow rate through all such rings. Lets model radius 'r' from 0 to 3 (the actual radius) and sum the volumes. We do this using the integration which is a sophisticated form of summing. Hence, the total volume flow rate \(V_{total}= \int_{0}^{3} 8(10-r^{2}) 2蟺r dr \)

04

Evaluate the integral

Solving this integral would give us the total volume flow rate. \(V_{total}= \int_{0}^{3} 16蟺r(10 - r^{2}) dr = 16蟺[(10/2)r^{2} - (1/4)r^{4}]_{0}^{3} = 16蟺 [(5*9) - (81/4)] = 16蟺 (45 - 20.25) = 16蟺 (24.75) = 1230蟺 cubic inches per second.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Definite Integral

The concept of a definite integral is like measuring the area under a curve on a graph. This is especially handy in calculus for finding the sum of infinite small parts. In our exercise concerning oil flow, we're essentially "adding up" all these small flows of oil through tiny cylindrical shells in the pipe from the center to the edge.

Mathematically, a definite integral is written as datadatadatadatadatadatadatadatadatadatadatadatadatadata\( \int_{a}^{b} f(x) \, dx \), which tells us to sum up values of the function \( f(x) \) from \( x = a \) to \( x = b \). In our case:

• \( a \) is 0 since we start from the center
• \( b \) is 3, the outer radius of the pipe
• \( f(x) \) is the flow rate function \( 16\pi r(10 - r^2) \)

When we compute this definite integral, we're getting the total flow rate through the pipe.

Volume Flow Rate

Volume flow rate, in simple terms, is how much fluid passes a point in a pipe over a certain time. Think of it as the speed and volume combined.

In our problem, the oil flow rate in the cylindrical pipe is influenced by the speed at which the oil moves and the cross-sectional area it moves through. The speed of the oil, given by the problem, reduces farther from the center, modeled as \(8(10-r^2)\). When you include the cross-sectional area (a thin ring) into this calculation, you end up with the expression for the volume flow rate. This is because:

• The speed gives you how fast the oil moves.
• The cross-sectional area tells you how much space the oil has to move through.

Thus, multiplying these gives the volume flow rate \(8(10-r^2) \times 2\pi r \Delta r\).

Cylindrical Shell Method

The Cylindrical Shell Method is a technique for finding the volume of a solid of revolution. It's especially useful when dealing with problems where slices are awkward to use.

In our exercise, this method allows us to think about the pipe as being composed of many 鈥渢hin rings鈥� or shells. Here's the main idea:

• Imagine peeling one layer off a cylinder and flattening it out. It resembles a rectangle.
• The height of each of these "rectangular" shells is based on how the flow rate changes, given by \(8(10-r^2)\).
• The circumference of the shell corresponds to the length of the rectangle, given by \(2\pi r\).

When we integrate these across the radius, we're summing the volumes of all these shells鈥攜ielding the total oil flow rate through the entire pipe.