Question

Comparing volumes Let \(R\) be the region bounded by \(y=\sin x\) and the \(x\) -axis on the interval \([0, \pi]\). Which is greater, the volume of the solid generated when \(R\) is revolved about the \(x\) -axis or the volume of the solid generated when \(R\) is revolved about the \(y\) -axis?

Short answer

Answer: The volume of the solid generated when R is revolved about the y-axis (Vy) is greater than the volume of the solid generated when R is revolved about the x-axis (Vx).

Step-by-step solution

Find the volume of the solid when R is revolved about the x-axis using the Disk Method

We will compute the volume by integrating the area of each disk from 0 to π along the x-axis. The disk area is given by π (radius)² where the radius is the function value (y = sin(x)). We have: \(V_x = \int_{0}^{\pi} \pi\big(\sin{x}\big)^2 dx\)

Evaluate the integral for Volume around the x-axis

To evaluate the integral, we can use the power reduction formula: \(\sin^2{x} = \frac{1-\cos{2x}}{2}\) \(V_x = \int_{0}^{\pi} \pi\left(\frac{1-\cos{2x}}{2}\right) dx\) Now we can distribute the pi and integrate: \(V_x = \pi\int_{0}^{\pi}\frac{1}{2} - \pi\int_{0}^{\pi} \frac{\cos{2x}}{2}dx\) Then, we have: \(V_x = \frac{\pi}{2}(x - \frac{\sin{2x}}{4}) \Big|_0^\pi\) Thus, \(V_x = \frac{\pi^2}{2}\)

Find the volume of the solid when R is revolved about the y-axis using the Disk Method

To find the volume when R is revolved about the y-axis, we must first express x in terms of y using the original function y = sin(x). This gives us x = arcsin(y). Now, using the disk method, we will compute the volume by integrating the area of each disk from 0 to 1 along the y-axis: \(V_y = \int_{0}^{1} \pi\left(\arcsin{y}\right)^2 dy\)

Evaluate the integral for Volume around the y-axis

This integral is a bit more complicated. We'll need to use substitution: Let \(u = \arcsin{y}\), then \(y = \sin{u}\) and \(dy = \cos{u}du\). Also, the new limits will be \(u=0\) and \(u=\pi/2\). The integral becomes: \(V_y = \int_{0}^{\frac{\pi}{2}} \pi\left(u\right)^2 \cos{u} du\) Now integrate by parts. Let \(v = u^2\) and \(dw = \cos{u} du\) then \(dv= 2udu\) and \(w=\sin{u}\): \(V_y = \pi \left[u^2\sin{u} \Big|_0^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} 2u\sin{u} du\right]\) Evaluating the first term, we have: \(V_y = \pi \left[\frac{\pi^2}{4}\sin{\frac{\pi}{2}}\right] - \pi\left[ \int_{0}^{\frac{\pi}{2}} 2u\sin{u} du\right]\) Now, we must integrate by parts again: Let \(v=2u\) and \(dw=\sin{u}du\) then \(dv=2du\) and \(w=-\cos{u}\): \( \int_{0}^{\frac{\pi}{2}} 2u\sin{u} du = -2u\cos{u}\Big|_0^{\frac{\pi}{2}} + \int_{0}^{\frac{\pi}{2}}2\cos{u}du = -2u\cos{u}\Big|_0^{\frac{\pi}{2}} + 2\sin{u}\Big|_0^{\frac{\pi}{2}}\) Thus, we have: \(V_y = \pi\left[\frac{\pi^2}{4} - \left(\left(\frac{\pi}{2}\right)\cos{\frac{\pi}{2}}+ 2\sin{\frac{\pi}{2}}\right)\right]\) Finally, \(V_y = \frac{\pi^3}{4}\)

Compare the volumes and determine which one is greater

Comparing the volumes: \(V_x = \frac{\pi^2}{2}\) and \(V_y = \frac{\pi^3}{4}\) Since \(\pi>2\), we can see that \(\frac{\pi^3}{4} > \frac{\pi^2}{2}\) Thus the volume of the solid generated when R is revolved about the y-axis (Vy) is greater than the volume of the solid generated when R is revolved about the x-axis (Vx).

Key concepts

Disk Method

The Disk Method is a technique used to find the volume of a solid of revolution. It's perfect for solids that are created when a region in a plane is revolved around an axis. For instance, when a curve like \(y = \sin x\) is revolved about the \(x\)-axis, each "disk" perpendicular to the \(x\)-axis becomes a circle. The radius of these disks is the function's value, \(y = \sin x\), at each \(x\).
To find the volume, we integrate the area of these circles across the interval of interest. The formula for the disk's area is \(\pi (\text{radius})^2\). Therefore, the volume \(V_x\) when revolving around the \(x\)-axis is:

• \(V_x = \int_0^{\pi} \pi(\sin x)^2 \, dx\)
• This formula arises by summing the infinitesimally small disk areas \(\pi (\sin x)^2\) over the interval from \(0\) to \(\pi\).

Integration by Parts

Integration by Parts is a powerful technique for finding integrals, especially useful when dealing with products of functions. The formula is derived from the product rule for differentiation and is stated as:

• \(\int u \, dv = uv - \int v \, du\)

This method shines in step-by-step solutions when encountering integrals like \(\int u^2 \cos u \, du\).
For the y-axis rotation in our problem, \(u = \arcsin(y)\), and the corresponding integral becomes \(\int \pi u^2 \cos u \, du\).
Using integration by parts allows us to break this down further:

• Select \(v = u^2\) and differentiate to find \(dv = 2u \, du\).
• Choose \(dw = \cos u \, du\) and integrate for \(w = \sin u\).

Each step simplifies the expression, leading to a more manageable integral to solve.

Power Reduction Formula

The Power Reduction Formula helps simplify integrals involving trigonometric powers. In this exercise, it is used to transform \(\sin^2{x}\) into a more manageable form:

• \(\sin^2{x} = \frac{1 - \cos{2x}}{2}\)

Why is this important? Integrating \(\sin^2{x}\) directly is complex, but using the formula reduces it to simpler terms involving \(\cos{2x}\).
This transformation allows for easier integration over the interval from \(0\) to \(\pi\).
Once reduced, the integration process for the volume when revolving around the \(x\)-axis becomes:

• \(V_x = \pi \int_0^{\pi} \left(\frac{1}{2} - \frac{\cos{2x}}{2}\right) \, dx\)

By breaking down the integral into more elementary terms, the Power Reduction Formula aids in handling trigonometric integrals efficiently.

Function Transformation

Function Transformation involves changing the form of a function to facilitate calculations. Particularly useful when the axis of revolution isn't the same as the variable of integration.
In the y-axis revolution for this problem, where \(y = \sin x\) is revolved about the \(y\)-axis, a transformation is necessary:

• The function \(y = \sin x\) must be rewritten as \(x = \arcsin(y)\).

This transformation shifts the focus from \(x\) to \(y\) and allows integrating with respect to \(y\).
For the disk method, this necessitates reevaluating the radius in terms of \(y\), leading to the integral:

• \(V_y = \int_0^1 \pi (\arcsin{y})^2 \, dy\)

Transforming functions in this way accommodates different axes of rotation and aids in calculating volumes otherwise challenging with the original function.