Question

A region of the Cartesian plane is described. Use the Shell Method to find the volume of the solid of revolution formed by rotating the region about each of the given axes. Region bounded by \(y=x^{2}-2 x+2\) and \(y=2 x-1\) Rotate about: (a) the \(y\) -axis (b) \(x=1\) (c) \(x=-1\)

Short answer

(a) \(\frac{5\pi}{2}\), (b) \(\frac{\pi}{3}\), (c) \(5\pi\).

Step-by-step solution

Find the Points of Intersection

To find the points of intersection of the curves, set the equations equal to each other: \[x^2 - 2x + 2 = 2x - 1\]Simplify to find the quadratic equation:\[x^2 - 4x + 3 = 0\]Factorize the quadratic equation:\[(x-3)(x-1) = 0\]Thus the points of intersection are at \(x = 1\) and \(x = 3\).

Formulate the Shell Method Integral for Part (a)

For rotation about the \(y\)-axis, the height of the shell is given by the difference in \(y\)-values of the functions: \[h(x) = (2x - 1) - (x^2 - 2x + 2) = -x^2 + 4x - 3\]The radius of the shell is \(x\). Therefore, the volume integral is:\[V = 2\pi \int_{1}^{3} x(-x^2 + 4x - 3) \, dx\]

Calculate the Integral for Part (a)

First, expand the integrand:\[-x^3 + 4x^2 - 3x\]Now, integrate each term individually:\[V = 2\pi \int_{1}^{3} (-x^3 + 4x^2 - 3x) \, dx\]This becomes:\[= 2\pi \left[ -\frac{x^4}{4} + \frac{4x^3}{3} - \frac{3x^2}{2} \right]_{1}^{3}\]Calculate the definite integral by substituting the limits of integration.

Compute the Limits for Part (a)

Substitute the upper limit (3) and lower limit (1):\[V = 2\pi \left( \left( -\frac{3^4}{4} + \frac{4 \times 3^3}{3} - \frac{3 \times 3^2}{2} \right) - \left( -\frac{1^4}{4} + \frac{4 \times 1^3}{3} - \frac{3 \times 1^2}{2} \right) \right)\]Simplify the expression:For \(x = 3\):\[-\frac{81}{4} + 36 - \frac{27}{2} = -\frac{9}{4}\]For \(x = 1\):\[-\frac{1}{4} + \frac{4}{3} - \frac{3}{2} = -\frac{11}{12}\]Then, the volume is:\[V = 2\pi \left( -\frac{9}{4} + \frac{11}{12} \right) = 2\pi \left( -\frac{15}{12} \right) = -\frac{5\pi}{2}\]However, since volume can't be negative, take the absolute value, giving final result:\[V = \frac{5\pi}{2}\]

Formulate the Shell Method Integral for Part (b)

For rotation about the line \(x=1\), the radius of the shell is \(|x-1|\) and the height remains as in Step 2. Thus, the volume integral is:\[V = 2\pi \int_{1}^{3} |x-1|(-x^2 + 4x - 3) \, dx\]Considering the interval and symmetry, we have:\[V = 2\pi \int_{1}^{3} (x-1)(-x^2 + 4x - 3) \, dx\]

Calculate the Integral for Part (b)

Distribute \((x-1)\) into the function:\[(-x^2 + 4x - 3)(x-1) = -x^3 + 5x^2 - 7x + 3\]Integrate this:\[V = 2\pi \int_{1}^{3} (-x^3 + 5x^2 - 7x + 3) \, dx\]Which gives:\[= 2\pi \left[ -\frac{x^4}{4} + \frac{5x^3}{3} - \frac{7x^2}{2} + 3x \right]_{1}^{3}\]Calculate the definite integral.

Compute the Limits for Part (b)

Substitute the limits in a similar manner as in Step 4:For \(x = 3\):\[-\frac{81}{4} + 45 - \frac{63}{2} + 9 = \frac{3}{4}\]For \(x = 1\):\[-\frac{1}{4} + \frac{5}{3} - \frac{7}{2} + 3 = \frac{11}{12}\]Thus, the volume is:\[V = 2\pi \left( \frac{3}{4} - \frac{11}{12} \right) = 2\pi \times \frac{1}{6} = \frac{\pi}{3}\]

Formulate the Shell Method Integral for Part (c)

For rotation about the line \(x=-1\), the radius of the shell is \(x+1\) and the height remains the same. Thus the volume integral is:\[V = 2\pi \int_{1}^{3} (x+1)(-x^2 + 4x - 3) \, dx\]

Calculate the Integral for Part (c)

Distribute \((x+1)\) into the function:\[(-x^2 + 4x - 3)(x+1) = -x^3 + 3x^2 + x - 3\]Integrate this:\[V = 2\pi \int_{1}^{3} (-x^3 + 3x^2 + x - 3) \, dx\]Which gives:\[= 2\pi \left[ -\frac{x^4}{4} + x^3 + \frac{x^2}{2} - 3x \right]_{1}^{3}\]Calculate the definite integral.

Compute the Limits for Part (c)

Substitute the limits as in previous steps:For \(x = 3\):\[-\frac{81}{4} + 27 + \frac{9}{2} - 9 = \frac{1}{4}\]For \(x = 1\):\[-\frac{1}{4} + 1 + \frac{1}{2} - 3 = -\frac{9}{4}\]Thus, the volume is:\[V = 2\pi \left( \frac{1}{4} + \frac{9}{4} \right) = 5\pi\]

Summary of Volumes

(a) Volume about the \(y\)-axis: \(\frac{5\pi}{2}\).(b) Volume about \(x=1\): \(\frac{\pi}{3}\).(c) Volume about \(x=-1\): \(5\pi\).

Key concepts

Volume of Solid of Revolution

When we talk about the volume of a solid of revolution, we are referring to the 3D shape created by rotating a 2D region around a straight line (axis of rotation). This concept is fundamental in calculus and very useful in many fields such as engineering and physics.

In our specific problem, the region enclosed by the curves is revolved around several axes (the y-axis, x = 1, and x = -1) to create different solids. The volume of the resulting solid can be calculated using integral calculus, which opens up quite a fascinating possibility of understanding how geometric shapes transform through rotation.

• The shell method is often advantageous over the disk method for certain axis rotations, specifically when the function is more naturally expressed as a height with respect to the axis of rotation.
• Typically used when the axis of rotation is vertical for functions y = f(x), which means the shell height is the difference between two functions y-values.

Knowing how to compute the volume of these solids helps in visualizing and practically applying mathematical operations to real-world problems involving rotational volumes.

Integral Calculus

Integral calculus is a significant area of calculus focusing on integrals and their properties. It involves summing up infinitesimal quantities to determine the whole, such as finding areas under curves, among others.

The shell method uses integral calculus to calculate the volume of solids. By setting up an integral that represents a small shell or slice's volume and then adding them together, you can find the total volume of the solid. This is where the power of integral calculus truly shines: it allows for precise calculations of complex shapes based on simple geometrical insights.

• The key element in integral calculus is the formation of an integrand, derived from the function being rotated.
• The shell method typically results in a volume integral of the form: \[V = 2\pi \int {radius \times height \, dx}\]

Understanding these fundamentals simplifies the complex process of creating these equations, guiding you to the solution effectively.

Definite Integral

A definite integral calculates the net area under a curve over a specific interval. It is denoted by the integral symbol with upper and lower limits representing the start and end of the interval.

In the context of finding the volume of solids of revolution through the shell method, definite integrals come into play significantly. The process involves determining the area enclosed between rotated curves across specified limits (intersection points in this context).

• The specific limits in this problem are given by the intersection points of the curves, from x = 1 to x = 3.
• Definite integrals ensure that the calculation remains bounded and provides a precise measure of volume.

Thus, definite integrals are crucial as they contain all necessary information to accurately calculate the volume once set up correctly with respect to the functions and limits provided.

Volume Integral

A volume integral specifically refers to the integral equation set up to calculate a volume, especially of solids formed by rotating curves. It transforms the conceptual understanding of rotation into a calculable form.

In the shell method, each shell can be seen as a cylindrical element. Calculating its volume involves multiplying its circumference, height, and thickness, expressed in terms of a volume integral. This involves:

• Radius of rotation: varies based on axis, simply the distance from the axis.
• Height of shell: difference in y-values depending on the function.
• Thickness: an infinitesimally small width, represented as dx.

For the rotational axes (y-axis, x = 1, x = -1), different volume integrals are set up to respect the symmetry and geometry of each resulting solid. Each integral provides detailed precision to capture the entire volume of these three-dimensional bodies.