Chapter 1: Problem 76
\(x=y^{2}\) and \(x=3 y\) rotated around the \(y\) -axis using the washer method
Short Answer
Expert verified
The volume is approximately \(32.4\pi\).
Step by step solution
01
Identify the Region of Rotation
The given functions are \(x = y^2\) and \(x = 3y\). We need to find the interval of \(y\) where these two curves intersect. To do this, set \(y^2 = 3y\), which gives \(y(y - 3) = 0\). Thus, \(y = 0\) and \(y = 3\) are points of intersection.
02
Set Up the Washer Method
When rotated about the \(y\)-axis, the disk washer method uses the formula: \(V = \pi\int_a^b(R^2 - r^2) \, dy\), where \(R\) is the outer radius and \(r\) is the inner radius. Here, \(a = 0\) and \(b = 3\) because those are the bounds of \(y\).Identify the outer and inner radii: \(R = 3y\) (outer radius), and \(r = y^2\) (inner radius).
03
Write the Integral Expression
Substitute the radii into the volume formula: \[ V = \pi \int_0^3 (3y)^2 - (y^2)^2 \, dy \]. This simplifies to \[ V = \pi \int_0^3 (9y^2 - y^4) \, dy \].
04
Evaluate the Integral
Compute the integral: \[ V = \pi \left[ \frac{9y^3}{3} - \frac{y^5}{5} \right]_0^3 \]. Simplify the expression inside the brackets: \(\frac{9y^3}{3} - \frac{y^5}{5} = 3y^3 - \frac{y^5}{5}\).
05
Apply the Limits
Substitute the limits into the expression: \[ V = \pi \left( 3(3)^3 - \frac{(3)^5}{5} \right) - \pi \left( 3(0)^3 - \frac{(0)^5}{5} \right) \]. Simplify to get: \[ V = \pi (81 - \frac{243}{5} ) \].
06
Simplify and Calculate
Calculate the volume: \[ V = \pi \left( 81 - 48.6 \right) \]. Thus, \[ V = \pi (32.4) \]. So, \(V \approx 32.4\pi \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Volume of Solids of Revolution
The concept of the volume of solids of revolution is a fascinating area in calculus, which deals with calculating volumes of 3-dimensional shapes formed by rotating a 2-dimensional area around a line, usually an axis. This technique enables us to understand the volume of complex shapes by transforming them into more manageable mathematical problems.
When a region in a plane is revolved around a line, like the y-axis, it creates a solid shape. For example, if you take the region between the curves of the functions, like a parabola and a line, and rotate it, you can form a doughnut-like shape, known as a toroid. The Washer Method is particularly useful for these situations where the shape has a hole in the center, distinguishing it from the Disk Method, which is used for solids without a central cavity. Understanding these methods lays the groundwork for more advanced applications in engineering, physics, and computer graphics.
When a region in a plane is revolved around a line, like the y-axis, it creates a solid shape. For example, if you take the region between the curves of the functions, like a parabola and a line, and rotate it, you can form a doughnut-like shape, known as a toroid. The Washer Method is particularly useful for these situations where the shape has a hole in the center, distinguishing it from the Disk Method, which is used for solids without a central cavity. Understanding these methods lays the groundwork for more advanced applications in engineering, physics, and computer graphics.
Integral Calculation
Integral calculation is the backbone of many calculus problems, especially when dealing with volumes of solids formed by rotation. In the Washer Method, the integral is used to sum up infinitely small slices of the solid to find the total volume. The general formula is represented as: \[ V = \pi \int_a^b \left( R^2 - r^2 \right) dy \]where \( R \) is the outer radius and \( r \) is the inner radius.
Here, the integral effectively accumulates the volume `slice by slice`, along the defined axis between two limits \( a \) and \( b \), which are found by investigating where the curves intersect. Understanding how to set up and simplify the integral and then evaluate it is crucial for successful application of the Washer Method. This requires careful attention to ensuring correct substitution and arithmetic simplification, which eventually leads to determining the exact volume of the revolved solid.
Here, the integral effectively accumulates the volume `slice by slice`, along the defined axis between two limits \( a \) and \( b \), which are found by investigating where the curves intersect. Understanding how to set up and simplify the integral and then evaluate it is crucial for successful application of the Washer Method. This requires careful attention to ensuring correct substitution and arithmetic simplification, which eventually leads to determining the exact volume of the revolved solid.
Intersection of Curves
Finding the intersection of curves is a vital step in solving problems with the Washer Method. This process identifies the bounds for the integral, which determines the region we are revolving to form our solid.
In our given example, the functions \( x = y^2 \) and \( x = 3y \) intersect. To find these intersection points, you need to solve the equation: \( y^2 = 3y \). By rearranging it to \( y(y - 3) = 0 \), we find solutions at \( y = 0 \) and \( y = 3 \). These values serve as the integration limits \( a \) and \( b \), marking the span of the solid along the \( y \)-axis.
Recognizing intersection points helps discern the exact region contributing to the volume and is essential for setting up the integral correctly. Without these, the calculation would lack the necessary boundaries and lead to incorrect results.
In our given example, the functions \( x = y^2 \) and \( x = 3y \) intersect. To find these intersection points, you need to solve the equation: \( y^2 = 3y \). By rearranging it to \( y(y - 3) = 0 \), we find solutions at \( y = 0 \) and \( y = 3 \). These values serve as the integration limits \( a \) and \( b \), marking the span of the solid along the \( y \)-axis.
Recognizing intersection points helps discern the exact region contributing to the volume and is essential for setting up the integral correctly. Without these, the calculation would lack the necessary boundaries and lead to incorrect results.
Outer and Inner Radii
In the Washer Method, understanding the roles of the outer and inner radii is crucial as they define the solid's structure. These radii are the horizontal distances from the axis of rotation, the y-axis in our example, to the curves that form the solid's walls.
For the problem at hand, the outer radius \( R \) is determined by the line furthest from the y-axis: \( R = 3y \). This is the function that extends further compared to the inner radius curve. Meanwhile, the inner radius \( r \) is marked by the curve closer to the axis: \( r = y^2 \). This setup means that the volume between the outer and inner radii is the area contributing to the solid's volume.
Understanding how to derive and use these radii is fundamental for applying the Washer Method. They are key in subtracting the inner volume from the outer volume within the integral, ultimately leading to the correct calculation of the solid's total volume.
For the problem at hand, the outer radius \( R \) is determined by the line furthest from the y-axis: \( R = 3y \). This is the function that extends further compared to the inner radius curve. Meanwhile, the inner radius \( r \) is marked by the curve closer to the axis: \( r = y^2 \). This setup means that the volume between the outer and inner radii is the area contributing to the solid's volume.
Understanding how to derive and use these radii is fundamental for applying the Washer Method. They are key in subtracting the inner volume from the outer volume within the integral, ultimately leading to the correct calculation of the solid's total volume.
