Question

Explain why or why not Determine whether the following statements are true and give an explanation or a counterexample. a. If the curve \(y=f(x)\) on the interval \([a, b]\) is revolved about the \(y\) -axis, the area of the surface generated is $$\int_{f(a)}^{f(b)} 2 \pi f(y) \sqrt{1+f^{\prime}(y)^{2}} d y$$ b. If \(f\) is not one-to-one on the interval \([a, b],\) then the area of the surface generated when the graph of \(f\) on \([a, b]\) is revolved about the \(x\) -axis is not defined. c. Let \(f(x)=12 x^{2} .\) The area of the surface generated when the graph of \(f\) on [-4,4] is revolved about the \(x\) -axis is twice the area of the surface generated when the graph of \(f\) on [0,4] is revolved about the \(x\) -axis. d. Let \(f(x)=12 x^{2} .\) The area of the surface generated when the graph of \(f\) on [-4,4] is revolved about the \(y\) -axis is twice the area of the surface generated when the graph of \(f\) on [0,4] is revolved about the \(y\) -axis.

Short answer

Question: Determine if the following statements about surface area generated by revolving a curve around an axis are true. If true, provide some explanation. If false, provide a counterexample. a) If a curve with equation \(y = f(x)\) is revolved about the y-axis, its surface area on the interval \([c,d]\) is given by $$S = 2\pi \int_{c}^{d} y \sqrt{1 + (\frac{d x}{d y})^{2}} dy$$ b) If a function is not one-to-one on a given interval, we cannot compute the surface area of the figure generated by revolving that curve around its x-axis. c) If a curve defined by the equation \(y = 12x^2\) is revolved around the x-axis, its surface area on the interval \([-4, 4]\) is twice its surface area generated on the interval \([0, 4]\). d) If a curve defined by the equation \(y = 12x^2\) is revolved around the y-axis, its surface area on the interval \([-4, 4]\) is twice its surface area generated on the interval \([0, 4]\). Answer: a) False. The integral limits should be over x-values when revolving around the y-axis. b) True. Revolving a curve that is not one-to-one around the x-axis can result in overlapping and an undefined surface area. c) True. The parabola defined by \(f(x) = 12x^2\) is symmetric with respect to the y-axis, so its surface area on the interval \([-4, 4]\) is twice its surface area on the interval \([0, 4]\) when revolved around the x-axis. d) False. The surface area when revolving around the y-axis differs when considering only the positive part of the curve versus the full curve, so the surface area on the interval \([-4, 4]\) is not twice the surface area on the interval \([0, 4]\).

Step-by-step solution

Statement a

Given the curve \(y = f(x)\), the interval \([a, b]\) is along the x-axis. If we rotate the curve about the y-axis, we would be using the formula for surface area of revolution about the y-axis: $$S = 2\pi \int_{a}^{b} x \sqrt{1 + (\frac{d y}{d x})^{2}} dx$$ However, the statement provides the formula with respect to "y" instead of "x". This is incorrect because the integral limits should be over x-values when revolving around the y-axis. Counterexample: Suppose \(f(x) = x^2\). Revolving this curve around the y-axis over the interval \([1,2]\), we should get surface area in terms of \(x\), not \(y\). The provided formula is incorrect, so statement a is false.

Statement b

If a function is not one-to-one on the interval \([a, b]\), it means that there are multiple y-values for an x-value on that interval. In this case, when revolving such a curve around the x-axis, it would result in overlapping and thus an undefined surface area. Therefore, statement b is true.

Statement c

Given \(f(x) = 12x^2\), we will find the surface area of revolution when revolving the graph of this function on the intervals \([-4, 4]\) and \([0, 4]\) around the x-axis and compare the two. The surface area of revolution formula around the x-axis is: $$S = 2\pi \int_{a}^{b} f(x) \sqrt{1 + (\frac{d f(x)}{d x})^{2}} dx$$ First we compute the derivative: \(f'(x) = 24x\). Now we compute surface area of revolution for the interval \([-4, 4]\) using formula above: $$S_{[-4,4]} = 2\pi \int_{-4}^{4} 12x^2 \sqrt{1 + (24x)^2} dx$$ Next, we compute surface area of revolution for the interval \([0, 4]\): $$S_{[0,4]} = 2\pi \int_{0}^{4} 12x^2 \sqrt{1 + (24x)^2} dx$$ Since the parabola defined by \(f(x) = 12x^2\) is symmetric with respect to the y-axis, \(S_{[-4,4]}\) should be exactly twice \(S_{[0,4]}\). Therefore, statement c is true.

Statement d

Given \(f(x) = 12x^2\), we will find the surface area of revolution when revolving the graph of this function on the intervals \([-4, 4]\) and \([0, 4]\) around the y-axis and compare the two. Remember that the surface area of revolution formula around the y-axis is: $$S = 2\pi \int_{a}^{b} x \sqrt{1 + (\frac{d y}{d x})^{2}} dx$$ First, express x in terms of y: \(x = \sqrt{\frac{y}{12}}\). Then, compute the derivative of x with respect to y: $$\frac{d x}{d y} = \frac{1}{24}y^{-\frac{1}{2}}$$ Now we compute surface area of revolution for the interval \([-4, 4]\) (in terms of x) using formula above: $$S_{[-4,4]} = 2\pi \int_{(-4)^{2}}^{(4)^{2}} \sqrt{\frac{y}{12}} \sqrt{1 + (\frac{1}{24}y^{-\frac{1}{2}})^2} dy$$ Next, we compute surface area of revolution for the interval \([0, 4]\): $$S_{[0,4]} = 2\pi \int_{0}^{(4)^{2}} \sqrt{\frac{y}{12}} \sqrt{1 + (\frac{1}{24}y^{-\frac{1}{2}})^2} dy$$ Notice that the surface area \(S_{[-4,4]}\) is not twice \(S_{[0,4]}\) because we have a different shape when revolving only the positive part of the curve around the y-axis, versus revolving the full curve (both positive and negative parts). Therefore statement d is false.

Key concepts

Integral Calculus

Integral Calculus is a branch of mathematics that focuses on the accumulation of quantities. One of its main tools is the concept of integration. Integration is used to find areas under curves, among other applications, and plays a crucial role in calculating the surface area of revolved curves.

When dealing with the surface area of a curve rotated around an axis, integration helps aggregate the infinitesimal surface elements created by the rotation. This is essential in defining and calculating exact surface areas, which would otherwise be challenging to compute using simple geometric tools.

Integral calculus allows us to consider small, manageable pieces of a curve, integrate them over the given interval, and thus gain an understanding of more complex surfaces generated by rotation. Integration becomes the key operator in surface area formulas, allowing for precise solution finding where algebraic methods and geometric estimation fail. By leveraging the power of calculus, we can explore not only basic shapes but also the surfaces of more complicated objects in mathematics.

Curve Rotation

Curve rotation is a fascinating concept in mathematics. It involves taking a curve and rotating it around a specified axis to create a three-dimensional surface. This operation is fundamental to finding the surface area of revolution.

The process of rotating a curve around an axis is akin to using the curve as a "template" that generates a symmetrical 3D shape. The axis of rotation is central to the shape that emerges. For example, when a curve is rotated around the y-axis, the profile of the curve dictates how the resultant object will look. The symmetry of the revolution ensures that each point on the curve contributes evenly to the shape's surface.

The importance of curve rotation extends beyond mathematics. It's a concept applied in engineering, physics, and computer graphics, among other fields. It helps visualize and design forms and structures by understanding the surfaces of rotation as created by simple 2D curves turned in space.

Surface Area Formula

The surface area formula is a method to calculate the total area that the surface of a shape encloses. When dealing with surfaces of revolution, this formula becomes a vital tool.

For a curve that is rotated around the x-axis, the formula used to find the surface area of the resulting shape is:

• \[ S = 2\pi \int_{a}^{b} f(x) \sqrt{1 + [f'(x)]^2} \, dx \]

This integral essentially sums up all the small circular rings (each with its own surface area) that make up the 3D surface of the shape.

Each small ring is determined by slicing the solid perpendicular to the axis of rotation. The more precise the slicing (i.e., the finer the integral), the more accurate the calculation of the surface area becomes. Understanding this formula allows one to handle a variety of tasks in mathematical modeling and design, where calculating areas accurately is crucial.

Symmetry in Calculus

Symmetry in calculus plays an important role when discussing the surface areas of revolved shapes. It greatly simplifies the calculation process by allowing mathematicians to exploit inherent regularities in functions or shapes.

In the context of surfaces of revolution, symmetry indicates that if a function or shape exhibits symmetry with respect to an axis, only a portion of it needs to be considered. Calculations performed on this portion can be used to deduce properties about the whole shape.

For example, the graph of a function such as

• \( f(x) = 12x^2 \)

is symmetric with respect to the y-axis. This implies that calculations over one interval can directly translate to the entire interval, reducing computational load and error.

By understanding symmetry, calculations can be efficiently structured around these axes or points of balance. It reduces complexities and complements the power of integral calculus by offering strategies to exploit structural uniformities.